Lesson: Electricity

Question 1

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio $\frac{R}{R'}$ is:

(a) $\frac{1}{25}$
(b) $\frac{1}{5}$
(c) 5
(d) 25

Solution:

Question 2

Which of the following terms does not represent electrical power in a circuit?

(a) $I^{2} R$
(b) $I R^{2}$
(c) $V I$
(d) $\frac{V^{2}}{R}$

Solution:

Question 3

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:

(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

Solution:

Question 4

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be:

(a) 1:2

(b) 2:1

(d) 4:1

Solution:

Question 5

How is a voltmeter connected in the circuit to measure the potential difference between two points?

Solution:

A voltmeter should be connected in parallel to measure the potential difference between two points.

Question 6

A copper wire has diameter 0.5 mm and resistivity of $1.6 \times 10^{- 8} Ω m$ . What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Solution:

Area of cross-section of the wire, $A = π {(\frac{D}{2})}^{2}$ .
Radius $= \frac{diameter}{2} = (\frac{0.5}{2}) mm = (\frac{0.0005}{2}) m$
Resistance, $R = 10 Ω$ .
The resistance is given by,

$R = ρ \frac{l}{A}$ . Therefore the length of the wire,

$l = \frac{R A}{ρ}$

$= \frac{10 \times 3.14 \times \frac{{(0.0005)}^{2}}{2^{2}}}{1.6 \times 10^{- 8}}$

$= \frac{10 \times 3.14 \times 25}{1.6 \times 4}$

$= 122.72 m$

If the diameter of the wire is doubled,

the new diameter $= 2 \times 0.5 = 1 mm = 0.001 m$

Therefore, the new resistance

$= ρ \frac{l}{A}$

$= \frac{1.6 \times 10^{- 8} \times 122.72 \times 4}{π \times {(\frac{1}{2} \times 10^{- 3})}^{2}}$

$= \frac{1.6 \times 10^{- 8} \times 122.72 \times 4}{3.14 \times 10^{- 5}}$

$= 250.2 \times 10^{- 2}$

$= 25 Ω$

Question 7

The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

I(amperes) 0.5, 1.0, 2.0, 3.0, 4.0

V (volts) 1.6, 3.4, 6.7, 10.2, 13.2

Plot a graph between V and I and calculate the resistance of that resistor.

Solution:

The coordinates for the graph is given below.

V (volts)	1.6	3.4	6.7	10.2	13.2
I (amperes)	0.5	1.0	2.0	3.0	4.0

The corresponding graph is given below

Resistance of the resistor:

The resistance of the resistor is equal to the slope of the line formed by the V-I graph.

So, the slope $= \frac{V}{I} = \frac{B C}{A C} = \frac{6.8}{2}$

$R = \frac{6.8}{2} = 3.4 Ω$ .

Therefore, the resistance is 3.4 Ω.

Question 8

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Solution:

Resistance of a resistor, $R = \frac{V}{I}$
Here the potential difference, V = 12 V and the current in the circuit,

$\begin{array}{l} I = 2.5 mA \\ = 2.5 x 10^{- 3} A \end{array}$

Therefore,

$\begin{array}{l} R = \frac{1.2}{2.5 \times 10^{- 3}} \\ = 4.8 \times 10^{3} Ω \end{array}$

Therefore, the resistance of the resistor is $4.8 \times 10^{3} Ω$ .

Question 9

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistors?

Solution:

Since all the resistors are connected in series, the current flow through all the component is the same. The equivalent resistance R

$\begin{array}{l} = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω \\ = 13.4 Ω . \end{array}$

As per Ohm’s law:

$\begin{array}{l} V = I R \\ \Rightarrow I = \frac{V}{R} \end{array}$

Potential difference, $V = 9 V$

$\Rightarrow I = \frac{9}{13.4} = 0.671 A$
As the current flow through each resistance will be the same in a series circuit, the current that will flow through the 12 Ω resistor is 0.671 A.

Question 10

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220V line?

Solution:

As per Ohm’s law V = IR
$\Rightarrow R = \frac{V}{I}$
Here, the supply voltage, $V = 220$ V and the current, $I = 5$ A
Let n be the number of resistors connected in parallel.

The equivalent resistance of the combination = R, is given as:

$\begin{array}{l} \frac{1}{R} = n \times \frac{1}{176} \\ \Rightarrow R = \frac{176}{n} \end{array}$

From Ohm’s law,

$\begin{array}{l} \frac{V}{I} = \frac{176}{n} \\ n = \frac{176 \times I}{V} \\ = \frac{176 \times 5}{220} \\ = 4 \end{array}$

Therefore, four resistors of 176 Ω are required to draw 5A of current on a 220 V line.

Question 11

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Solution:

a) Connecting all the resistors in series:

The equivalent resistance $= 6 Ω + 6 Ω + 6 Ω = 18 Ω$

b) Connecting all the resistors in parallel:

The equivalent resistance will be $= 2 Ω$

As both the results are not desired, a combination of series and parallel

connections would be required.

c) Connecting two resistors in parallel:

When two 6 Ω resistors are connected in parallel. Their equivalent resistance is:

$\begin{array}{l} = \frac{1}{\frac{1}{6} + \frac{1}{6}} \\ = \frac{6 \times 6}{6 + 6} \\ = 3 Ω \end{array}$

Connecting the third resistor in series with 3 Ω, we get the equivalent resistance $= 6 Ω + 3 Ω = 9 Ω$

d) Connecting two resistors in series:

When two 6 Ω resistors are in series, the equivalent resistance will be their sum, $6 + 6 = 12 Ω$

Connecting the third resistor in parallel with 12 Ω, we get the equivalent resistance as:

$\begin{array}{l} = \frac{1}{\frac{1}{12} + \frac{1}{6}} \\ = \frac{12 \times 6}{12 + 6} \\ = 4 Ω \end{array}$

The combination of resistors, as done in the previous two cases, give the desired resistance.

Question 12

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Solution:

Given:
Supply voltage, $V = 220$ V
Maximum allowable current, $I = 5$ A
Rating of an electric bulb, $P = 10$ watt.

Let R₁be the resistance of each lamp.
As the resistance, $R = \frac{V^{2}}{P}$

$\Rightarrow R_{1} = \frac{{(220)}^{2}}{10} = 4840 Ω$

According to Ohm’s law,

$V = I R$

Let R be the total resistance of the circuit for n number of electric bulbs

$\Rightarrow R = \frac{V}{I} = \frac{220}{5} = 44 Ω$

Since the resistance of each electric bulb, $R_{1} = 4840 Ω$

$\begin{array}{l} \Rightarrow \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + ..... upto n times \\ \Rightarrow \frac{1}{R} = \frac{1}{R_{1}} \times n \\ \Rightarrow n = \frac{R_{1}}{R} \\ = \frac{4840}{44} \\ = 110 \end{array}$

Therefore, the number of electric bulbs connected in parallel are 110.

Question 13

A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Solution:

Given:

The supply voltage, $V = 220 V$
The resistance of one coil, $R = 24 Ω$

Case 1: When the coils are used separately

Let I₁ be the current flowing through the coil.

According to Ohm's law,
$V = I_{1} R_{1}$

$\Rightarrow I_{1} = \frac{V}{R_{1}} = \frac{220}{24} = 9.166 A$

Thus,9.16 A current will flow through the coil when used separately.

Case 2: When the coils are connected in series

Let the total resistance of the circuit be R₂and the total current flowing be I_2.

The total resistance, $R_{2} = 24 Ω + 24 Ω = 48 Ω$

According to Ohm's law, $V = I_{2} R_{2}$
$\Rightarrow I_{2}_{} = \frac{V}{R_{2}} = \frac{220}{48} = 4.58 A$
Thus, when the coils are connected in series, 4.58A current will flow through the circuit.

Case 3: When the coils are connected in parallel

Let the total resistance of the circuit be R₃ and the current through it be I₃.

The total resistance, R₃ is given as

$\frac{1}{\frac{1}{24} + \frac{1}{24}} = \frac{24}{2} = 12 Ω$

According to Ohm's law,

$V = I_{3} R_{3}$

$\Rightarrow I_{3} = \frac{V}{R_{3}} = \frac{220}{12} = 18.33 A$

Thus, 18.33 A current will flow through the circuit.

Question 14

Compare the power used in the 2 Ω resistor in each of the following circuits:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistor, and

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Solution:

Case (i)

Given:

· Potential difference, $V = 6$ V

· 1 Ω and 2 Ω resistors are connected in series.

The equivalent resistance of the circuit, $R = 1 + 2 = 3 Ω$

Let I be the current through the circuit.
According to Ohm’s law,
$\begin{array}{l} V = I R \\ \Rightarrow I = \frac{6}{3} = 2 A \end{array}$

So, the current flowing through the 2 Ω resistor is 2 A.

The power used $= {(I)}^{2} R = {(2)}^{2} \times 2 = 8 W$

Case (ii)

Given:

· Potential difference, $V = 4$ V

· 12 Ω and 2 Ω resistors are connected in parallel.

Since the resistors are connected in parallel, the voltage across 2 Ω resistor is 4 V.

The power consumed by the resistor $= \frac{V^{2}}{R} = \frac{4^{2}}{2} = 8 W$

Question 15

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Solution:

Since both the bulbs are connected in parallel, the potential difference across each of them is same and is equal to 220 V.

The current drawn by the bulb of rating 60 W $= \frac{power}{voltage} = \frac{60}{220} A$ .

The current drawn by the bulb of rating 100 W $= \frac{power}{voltage} = \frac{100}{220} A$

Therefore, the total current drawn $= \frac{60}{220} + \frac{100}{220} = 0.727 A$

Question 16

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Solution:

If P is the power of the appliance and t is the time for which it is in use,

then, the energy consumed by the electrical appliance H= Pt.
Therefore, the energy consumed by the TV set of power 250 W in 1 h

$\begin{array}{l} = 250 \times 3600 \\ = 9 \times 10^{5} J \end{array}$

The energy consumed by the toaster of power 1200 W in 10 minutes

$\begin{array}{l} = 1200 \times 600 \\ = 7.2 \times 10^{5} J \end{array}$

Thus, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

Question 17

An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Solution:

If I is the current flowing through the circuit and R is the resistance of the circuit, the rate of heat produced by a device is the expression for power of the device which is equal to $I^{2} R$ .

Given:

The resistance of the electric heater, $R = 8 Ω$
The current drawn, $I = 15 A$

Therefore, the power, $P = {(15)}^{2} \times 8 = 1800 J/s$

Thus, the heat produced by the heater is at the rate of 1800 J/s.

Question 18

Explain the following:

(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and

electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity

transmission?

Solution:

(a) Tungsten is an alloy. It has a very high melting point and a very high resistivity. Thus, it offers very high resistance when current flows through it. Due to the high resistance, the tungsten becomes very hot and glows.

(b) The heating element of the heater is made up of alloy. An alloy offers very high resistance when current flows through it. Due to the high resistance, the alloy becomes very hot and glows. The heating elements are not made up of pure metal because metals are good conductor of electricity. Thus, they offer very little resistance in flow of electric current. Very little heat is produced when current passes through it.

i) In a series connection, the overall resistance of the circuit becomes high due to which the current supply from the power source is very low.

ii) If one appliance in series connection stops working, the other appliances too stop working. Similarly, if we have to use only appliance, the other appliances will also start working.

iii) All the appliances connected in series connection do not get the same voltage as supplied by power supply.

(d) When the area of cross section increases, the resistance decreases, and vice versa. Thus, the resistance (R) of a wire is inversely proportional to its area of cross-section (A).

(e) Copper and aluminium are good conductors of electricity. Thus, during transmission, the loss of current would be least. That is the reason why copper and aluminium are usually employed for electricity transmission.

Lesson: Electricity

Question 1

A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams figure given below. The current recorded in the ammeter will be:

(a) Maximum in (i)

(b) Maximum in (ii)

(d) The same in all the cases

Solution:

Question 2

In the following circuits, heat produced in the resistor or combination of resistors connected to a 12 V battery will be:

(a) Same in all the cases

(b) Minimum in case (i)

(d) Maximum in case (iii)

Solution:

Question 3

Electrical resistivity of a given metallic wire depends upon

(a) Its length

(b) Its thickness

(d) Nature of the material

Solution:

Question 4

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly:

(a) $10^{20}$

(b) $10^{16}$

(d) $10^{23}$

Solution:

Question 5

In the following circuits, identify the one in which the electrical components have been properly connected.

(a) (i)

(b) (ii)

(d) (iv)

Solution:

Question 6

What is the maximum resistance which can be made using five resistors each of $\frac{1}{5} Ω$ ?

(a) $\frac{1}{5} Ω$

(b) 10 Ω

(d) 1 Ω

Solution:

Question 7

What is the minimum resistance which can be made using five resistors each of $\frac{1}{5} Ω$ ?

(a) $\frac{1}{5} Ω$

(b) $\frac{1}{25} Ω$

(d) 25 Ω

Solution:

Question 8

In the given below figure, the proper representation of series combination of cells obtaining maximum potential is:

(a) (i)

(b) (ii)

(d) (iv)

Solution:

Question 9

Which of the following represents voltage?

(a) $\frac{Work done}{Current × Time}$

(b) $Work done × Charge$

(d) $Work done × Charge × Time$

Solution:

Question 10

A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section:

(a) $\frac{A}{2}$

(b) $\frac{3 A}{2}$

(d) 3 A

Solution:

Question 11

A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R₁, R₂ and R₃ respectively. Which of the following is true?

(a) $R_{1} = R_{2} = R_{3}$

(b) $R_{1} > R_{2} > R_{3}$

(d) $R_{2} > R_{3} > R_{1}$

Solution:

Question 12

If the current I through a resistor is increased by100% (assume that temperature remains unchanged), the increase in power dissipated will be:

(a) 100 %

(b) 200 %

(d) 400 %

Solution:

Question 13

The resistivity does not change if:

(a) The material is changed

(b) The temperature is changed

(d) Both material and temperature are changed

Solution:

Question 14

In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

(a) Brightness of all the bulbs will be the same

(b) Brightness of bulb A will be the maximum

(d) Brightness of bulb C will be less than that of B

Solution:

Question 15

In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be:

(a) 5 J

(b) 10 J

(d) 30 J

Solution:

Question 16

An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

(a) 1 A

(b) 2 A

(d) 5 A

Solution:

Question 17

Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have:

(a) Same current flowing through them when connected in parallel

(b) Same current flowing through them when connected in series

(d) Different potential difference across them when connected in parallel

Solution:

Question 18

Unit of electric power may also be expressed as:

(a) Volt ampere

(b) Kilowatt hour

(d) Joule second

Solution:

Short Answer Questions

Question 19

A child has drawn the electric circuit to study Ohm’s law as shown in given below figure. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

Solution:

Question 20

Three 2 Ω resistors, A, B and C, are connected as shown in the following figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting.

Find the maximum current that can flow through the three resistors?

Solution:

Maximum current through resistor A $= \sqrt{\frac{18}{2}} A = 3 A$

Given that the resistance of each of the resistors are 2 Ω each and maximum power the resistors can withstand is 18 W.

The power dissipated is given by, $P = I^{2} R$ .

The maximum current that the resistor A can withstand $= \sqrt{\frac{18}{2}} A = 3 A$

Since B and C are connected in parallel, the 3A of current that passes through the resistor A splits and passes through the resistor B and C equally. Thus, the maximum current through resistors B and C each,

$= 3 \times \frac{1}{2} A = 1.5 A$ .

Question 21

Should the resistance of an ammeter be low or high? Give reason.

Solution:

An ammeter measures the current passing through a circuit. Ideally, an ammeter should not have any resistance as resistance of the ammeter reduces the amount of current passing through it and therefore the actual amount of current passing through a circuit cannot be determined. However, in practice, every ammeter will have some amount of resistance at least. Therefore, the best option would be to have an ammeter with minimum possible resistance.

Question 22

Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4 Ω resistors? Give reason.

Solution:

When two resistors are connected in series, the potential difference across resistors will be equal only if the resistance of the resistors are equal.

The combined resistance of two resistors connected in parallel is 2 Ω.

Thus, the potential difference across the 2 Ω resistor will be the same as that across the parallel combination of two 4 Ω resistors.

Question 23

How does use of a fuse wire protect electrical appliances?

Solution:

A fuse consists of a piece of wire made up of an alloy or metal of an appropriate melting point. When the current higher than a specified value flows through a wire, the temperature of the fuse wire increases; leading to the melting and breakage of the circuit.

Thus, a fuse stops the flow of any unduly high electric current in a circuit. This helps in protecting electrical appliances from damages.

Question 24

What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Solution:

Resistivity is a measure of resistance offered to an electric current by a conducting substance.

Resistance of a conductor is given by the relation, $R = ρ \frac{l}{A}$

where l is the length, A is the area of cross-section of the wire and $ρ$ is resistivity of the wire.

In simple words, the more is the resistivity of a substance, the more is its resistance.

The SI unit of electrical resistivity is ohm-metre (Ω.m).

When the length of the wire is doubled, the resistance offered by the wire also doubles. This reduces the amount of current by half. This can be understood from the equation $I = \frac{V}{R}$ . Here I is the current and R is the resistance in a circuit across a voltage of V.

Question 25

What is the commercial unit of electrical energy? Represent it in terms of joules.

Solution:

The commercial unit of electrical energy is kWh.

$1 kWh = 1000 W \times 60 \times 60 s = 3.6 \times 10^{6} J$

Question 26

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp.

Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductors and potential difference across the lamp will take place? Give reason.

Solution:

Case 1: When the lamp is in a series circuit,

the total resistance of the circuit, $R = \frac{V}{I}$ ,

$R = \frac{10 V}{1 A} = 10 Ω$

The resistance offered by the lamp

$= total resistance of the circuit and the lamp - resistance of the circuit$

$= 10 Ω - 5 Ω$

$= 5 Ω$

Case 2: When a 10 Ω resistor is connected in parallel with the series connection, there will be no change in current flowing through 5 Ω conductors, as there will be no change in potential difference across the lamp either.

Question 27

Why is parallel arrangement used in domestic wiring?

Solution:

(i) In a parallel connection, the overall resistance of the circuit becomes less due to which the current supply from the power source is high.

(ii) Unlike series connection, if one appliance in parallel connection stops working, the other appliances are not affected. Similarly, if we have to use only appliance, we can do so independently.

(iii) All the appliances connected in parallel connection can get the same voltage as supplied by power supply.

Question 28

$B_{1} {, B}_{2} {and B}_{3}$ are three identical bulbs connected as shown in given below figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb B₁ gets fused?

(ii) What happens to the reading of $A_{1} {, A}_{2} {, A}_{3}$ and A when the bulb B₂ gets fused?

(iv) How much power is dissipated in the circuit when all the three bulbs glow together?

Solution:

(i) The glow of the bulbs B₂ and B₃ will remain the same because glow of bulbs depends on power. Power is given as $P = \frac{V^{2}}{R}$
The potential difference (V) and resistance (R) of B₂ and B₃remain the same. Therefore, there will be no change in the glow of b.

(ii) The amount of current flowing through each of the bulbs remain the same as the current, $I = \frac{V}{R}$ . Both V and R in this case remains the same.

(iii) The net resistance (R') of the three resistors connected in parallel can be found from the following equations.

$\frac{1}{R'} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} = \frac{3}{R}$

as $R_{1} = R_{2} = R_{3} = R$

Therefore,

$\begin{array}{l} R' = \frac{R}{3} \\ \Rightarrow V = I R \\ \Rightarrow 4.5 = 3 \times (\frac{R'}{3}) \\ \Rightarrow R' = 4.5 Ω \\ P = V \times I = 4.5 \times 3 = 13.5 W \end{array}$

Long Answer Questions

Question 29

Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.

(a) Will the bulb in the two circuits glow with the same brightness?

Justify your answer.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

Solution:

(a) No. The combined resistance of the bulbs in series will be three times the resistance of the bulbs connected in parallel. Therefore, the current in the series combination will be one-third of the current in each bulb in the parallel combination. Therefore, the bulbs in the parallel connections will glow brighter.

(b) When one of the bulbs in both the connections gets fused:

· The remaining bulbs connected in the series stop glowing as the circuit is broken.

· There will be no impact on the two bulbs as the circuits for these bulbs remain intact without any change in the voltage and the amount of current flowing through the circuits.

Question 30

State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Solution:

Ohm’s law states that the potential difference across the ends of a resistor is directly proportional to the current passing through the resistor provided its temperature remains the same. If I is the current flowing through a conductor, and V is the potential difference at its ends, then $V \propto I \Rightarrow V = I R$

where R is the constant of proportionality and commonly called the resistance of a conductor.

Experiment to verify Ohm’s Law:

Steps:

a) We can set up a circuit as shown in the figure. The circuit consists of a nichrome wire of length, say 1 m, an ammeter, a voltmeter and four cells of 2.5 V each.

b) We will use only one cell as the source in the circuit. We will note down the reading in the ammeter and the voltmeter for the potential difference across the nichrome wire in the circuit and tabulate these values.

c) Next, we will connect two cells in the circuit and note the respective readings of the ammeter and voltmeter.

d) We will repeat the above steps using three cells and then four cells in the circuit, separately.

e) From the table, we can calculate the value of the V and I for each observation.

S.No

Number of cells used in the circuit

Current through the nichrome wire, I (ampere)

Potential difference across the nichrome wire, V (volt)

$\frac{V}{I}$

( $\frac{Volt}{Ampere}$ )

f) We will then plot a graph between V and I by taking V along X-axis.

Observation: The V $-$ I graph is a straight line that passes through the origin of the graph.

Conclusion- $\frac{V}{I}$ is a constant ratio. This verifies Ohm’s Law.

This law does not hold good for all situations. For example:

· This law is valid only for conductors, provided the temperature and other physical conditions remain constant.

· It is not followed in case of insulators.

Question 31

What is electrical resistivity of a material? What is its unit?

Describe an experiment to study the factors on which the resistance of conducting wire depends.

Solution:

Resistivity is numerically equal to the resistance of a wire which is 1 metre long having a cross section of 1 square metre.

In simple words, the more is the resistivity of a substance, the more is its resistance per unit length and per unit area.

The SI unit of electrical resistivity is ohm-metre (Ω $�$ m).

Factors that affect the resistance of a wire:

a) Material (resistivity)

b) Cross- sectional area

c) Length

d) Temperature

Experiments:

a) Let us take wires of different materials like copper, aluminium, iron of same

cross-sectional area and of the same length.

b) Let us connect to a circuit as shown in the figure. Here AA’ represents a wire.

(i) Let us connect each wire one by one between A and A’. After that, let us insert

the key into the plug and note down the reading of the ammeter.

Observation: The value of the current is different for different wires.

Conclusion: Since the same cell is used every time, the potential difference (V) across the wires is same. This means different wires used in this experiment draw different currents when the same potential difference is applied across them.

Hence, R α ρ

(ii) Let’s perform the same activity as done earlier by taking the same wire of different cross sections.

Observation: The value of the current is different for different cross sections of the same wire. The more is the cross section, the more is the current flowing through the wire.

Conclusion: Resistance is inversely proportional to the area of cross-section (A)

$R α \frac{1}{A}$

(iii) Let’s perform the same activity as done before by taking the same wire of different lengths.

Observation: As the length increases, the current in the circuit decreases.

Conclusion: The resistance is proportional to length.

(iv) Finally, let’s perform the same experiment by increasing the temperature of the lab where the experiment is being performed.

Observation: The higher the temperature, the lesser is the current.

Conclusion: The higher the temperature, the higher is the resistance.

Question 32

How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Solution:

We can use three bulbs as resistors to perform this activity. We can make the connections as shown.

We will take readings of the ammeter by positioning it before the resistor $R_{1}$ , after the resistor $R_{3}$ , between $R_{1} and R_{2} and between R_{2} and R_{3}$ .

Observation: The ammeter reading for the given setup remains the same for every position.

Conclusion: The same current flows through every part of the circuit containing three resistances in series.

Question 33

How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Solution:

This can be proved by a very simple logic. Let us take a parallel connection of three resistors are shown. The potential difference across the resistor R₁ is same as the difference in the electric potential between the point X and Y.

Similarly, the potential difference across the resistors $R_{2} and R_{3}$ are same as the difference in the electric potential between the points X and Y.

Thus, we can conclude that the same potential difference (voltage) exists across the three resistors connected in a parallel arrangement to a battery.

Question 34

What is Joule’s heating effect? List its four applications in daily life.

Solution:

The heating of resistor because of dissipation of electrical energy is commonly known as “Heating Effect of Electric Current”. The heat lost in the process is explained by the Joule’s Law of Heating. The law states that the heat produced in a resistor is:

(i) Directly proportional to the square of current for a given resistance

(ii) Directly proportional to resistance for a given current, and

(iii) Directly proportional to the time for which the current flows through the resistor.

Four applications of Joule’s law of heating are:

a) Electrical appliances such as electric iron, electric toaster, etc.

b) Filaments of the electric bulbs,

c) Utilised in electric fuse for protection of household wiring and electric appliances,

d) Electric heater, room heater, geyser, etc.

Question 35

Find out the following in the electric circuit given in following figure.

(a) Effective resistance of two 8 Ω resistors in the combination

(b) Current flowing through 4 Ω resistor

(d) Power dissipated in 4 Ω resistor

(e) Difference in ammeter readings, if any

Solution:

(a) The two given resistors of 8 Ω each are connected in parallel.

For 2 resistors connected in parallel, the effective resistance R is given by:

$\begin{array}{l} \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\ R = \frac{R_{1} R_{2}}{R_{1} + R_{2}} \\ \Rightarrow R = \frac{8 \times 8}{8 + 8} \\ \Rightarrow R = \frac{64}{16} \\ \Rightarrow R = 4 Ω \end{array}$

(b) The current (I) flowing through the 4Ω resistor $= \frac{V}{R} = \frac{8}{4 + 4} = 1 A$ .

(d) The power dissipated in 4 Ω resistor $= I^{2} R = 1^{2} \times 4 = 4 W$ .

(e) Both the ammeters will show the same current reading as the same current

passes through each of these.

Lesson: Electricity

Question 1

What does an electric circuit mean?

Solution:

An electric circuit is a continuous and closed path through which electric current flows. It consists of electric devices like electric bulbs, fans, etc., a source of electricity, switches and wires that are connected.

Question 2

Define the unit of current.

Solution:

The unit of electric current is ampere (A).

1 A is the flow of 1 C of charge in 1 second through a wire.

i.e., $1 A= \frac{1 C}{1 s}$

Question 3

Calculate the number of electrons constituting one coulomb of charge.

Solution:

An electron possesses a charge of $1.6 \times 10^{- 19} C$ .

Therefore, the number of electrons that contain 1 C of charge

$\begin{array}{l} \begin{array}{l} = \frac{1}{1.6} \times 10^{- 19} \\ = 6.25 \times 10^{18} \end{array} \\ = 6 \times 10^{18} \end{array}$

Question 4

Name a device that helps to maintain a potential difference across a conductor.

Solution:

Devices like a battery, cell, power supply, etc., which are sources of electricity, help to maintain a potential difference across a conductor.

Question 5

What is meant by saying that the potential difference between two points is 1 V?

Solution:

When 1 joule of work is done to move a charge of 1 coulomb from one of the points to the other, the potential difference between those two points in a current carrying conductor is said to be 1 V.

i.e., $1 V= \frac{1 J}{1 C}$

Question 6

How much energy is given to each coulomb of charge passing through a 6V battery?

Solution:

The energy given to each coulomb of charge is equal to the amount of work done in moving it through a 6V battery.
Work done = potential difference × charge
Given that the charge = 1 C and the potential difference = 6V

$�$ Work done $= 6 \times 1 = 6 J$

Question 7

On what factors does the resistance of a conductor depend?

Solution:

Factors on which the resistance of a wire depends:

· Material used to make the conductor

· Length of the conductor

· Cross-sectional area of the conductor

· Temperature of the conductor

Question 8

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Solution:

The resistance to the flow of current in a conductor is inversely proportional to the area of its cross-section. A thick wire will have more area of the cross section. Thus, current will flow more easily through a thick wire than a thin wire of the same material, when connected to the same source.

Question 9

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Solution:

According to Ohm’s law,

$\begin{array}{l} V = I R \\ \Rightarrow I = \frac{V}{R} ... (1) \end{array}$

When the potential difference is decreased to half, the new potential difference will be $\frac{V}{2}$ .

As the resistance remains constant,

So, the new current flowing will be $= \frac{\frac{V}{2}}{R}$

$\begin{array}{l} = \frac{1}{2} \times \frac{V}{R} \\ = \frac{1}{2} \times I \\ = \frac{I}{2} \end{array}$

Thus, the amount of current flowing through an electrical component is reduced by half if the potential difference is decreased to half.

Question 10

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Solution:

Electrical appliances like electric toasters and electric irons work on the principle of heating effects of current. The coils of electric toasters and electric irons are made of an alloy rather than a pure metal because:

(a) The resistivity of an alloy is higher than the pure metal.

(b) The alloys do not melt readily at high temperature.

Question11

Use the data in the table below to answer the questions.

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

Solution:

(a) Resistivity of iron $= 10.0 \times 10^{- 8} Ω$
Resistivity of mercury $= 94.0 \times 10^{- 8} Ω$
Thus, iron is a better conductor than mercury.

(b) Silver is the best conductor as its resistivity is the lowest among the listed materials.

Question12

Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Solution:

Question13

Redraw the circuit of above question, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Solution:

The ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor 12 Ω, the voltmeter should be connected in parallel.

The reading of the ammeter:

Given:
Potential difference, $V = 6 V$
Resistance of the circuit, $R = 5 + 8 + 12 = 25 Ω$

According to Ohm’s law,

$\begin{array}{l} V = I R \\ \Rightarrow I = \frac{V}{R} = \frac{6}{25} = 0.24 A \end{array}$

Thus, the reading of the ammeter will be 0.24 A.

The reading of the voltmeter:

Let the potential difference across 12 Ω resistor be $V_{1}$
Current flowing through the 12 Ω resistor, $I = 0.24 A$
Using Ohm’s law, we get $V_{1} = I R = 0.24 \times 12 = 2.88 V$
This the reading of the voltmeter will be 2.88 V

Question14

Judge the equivalent resistance when the following are connected in parallel $-$

(a) 1 Ω and $10^{6}$ Ω,

(b) 1 Ω and $10^{3}$ Ω, and $10^{6}$ Ω.

Solution:

(a) When 1Ω and $10^{6}$ Ω are connected in parallel, the equivalent resistance R can be found as:

$\begin{array}{l} \frac{1}{R} = \frac{1}{1} + \frac{1}{10^{6}} \\ \Rightarrow R = \frac{10^{6}}{1 + 10^{6}} = \frac{10^{6}}{10^{6}} \approx 1 Ω \end{array}$

(b) When 1Ω, $10^{3}$ Ω and $10^{6}$ Ω are connected in parallel, the equivalent resistance R can be found by:

$\begin{array}{l} \Rightarrow \frac{1}{R} = \frac{1}{1} + \frac{1}{10^{3}} + \frac{1}{10^{6}} = \frac{10^{6} + 10^{3} + 1}{10^{6}} \\ \Rightarrow R = \frac{1000000}{1001001} = 0.999 Ω \end{array}$

Question15

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source.

What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Solution:

Given:

· Resistance of the electric lamp, $R_{1} = 100 Ω$

· Resistance of the toaster, $R_{2} = 50 Ω$

· Resistance of the water filter, $R_{3} = 500 Ω$

· Potential difference of the source, $V = 220 V$
As these are connected in parallel, the corresponding circuit diagram is given as:

Let R be the equivalent resistance of the circuit.

$\begin{array}{l} \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} \\ = \frac{5}{500} + \frac{10}{500} + \frac{1}{500} \\ = \frac{16}{500} \end{array}$

As per Ohm’s law,

$\begin{array}{l} V = I R \\ \Rightarrow I = \frac{V}{R} \\ \Rightarrow I = \frac{220}{\frac{500}{16}} = \frac{220 \times 16}{500} = 7.04 A \end{array}$

So, 7.04 A of current is drawn by all the three given appliances.

$\Rightarrow$ The current drawn by the electric iron connected to the same source of potential $220 V = 7.04 A$

If $R'$ is the resistance of electric iron then $V = I R'$

$R'$ $= \frac{V}{I} = \frac{220}{7.04} = 31.25 Ω$

Question16

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Solution:

The advantages of connecting electrical devices in parallel are:

a) The potential difference across all appliances connected in parallel is equal to the supplied voltage. Hence, there is no division of voltage among the appliances.

b) The total effective resistance of the circuit when connected in parallel is reduced. This results in more flow of current for a given voltage.

Question17

How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Solution:

(a) To get 4 Ω, we can make the connections as shown in the circuit diagram.

The equivalent resistance of 6 Ω and 3 Ω when connected in parallel

$= \frac{1}{\frac{1}{6} + \frac{1}{3}} = \frac{6 \times 3}{6 + 3} = 2 Ω$

Therefore, the equivalent resistance of the circuit 2 Ω connected with 2 Ω in series is 4 Ω.
(b) To get 1 Ω, we can make the connections as shown in the circuit diagram.

Here, the resistors are connected in parallel.

The equivalent resistance $= \frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{6}} = \frac{1}{\frac{3 + 2 + 1}{6}} = \frac{6}{6} = 1 Ω$

Question18 What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Solution:

There are four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω respectively.

(a) The highest resistance can be secured by connecting these coils in series.

The resultant resistance when connected in series is $4 + 8 + 12 + 24 = 48 Ω$ .
(b) The lowest resistance can be secured by connecting these coils in parallel. The resultant resistance is

$\frac{1}{\frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}} = \frac{1}{\frac{6 + 3 + 2 + 1}{24}} = \frac{24}{12} = 2 Ω$

Question19

Why does the cord of an electric heater not glow while the heating element does?

Solution:

A cord of an electric heater is usually made up of aluminium or copper. These are good conductors of electricity. Thus, they offer very little resistance to the flow of electric current.

The heating element of the heater is made up of an alloy that has a very high resistance. Thus, it offers very high resistance when current flows through it. Due to the high resistance, the alloy becomes very hot and glows.

Question20

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Solution:

Given:

Charge, $Q = 96000 C$
Time, $t = 1 hr = 60 \times 60 = 3600 s$

Potential difference, $V = 50 volts$
As $I = \frac{Q}{t}$

$\Rightarrow I = \frac{96000}{3600} = \frac{80}{3} A$

As $H = V I t$

$\Rightarrow H = 50 \times \frac{80}{3} \times 60 \times 60 = 4.8 \times 10^{6} J$

Therefore, the heat generated is 4.8 x 10⁶ J.

Question21

An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Solution:

The amount of heat (H) produced is given by the equation, $H = V l t$
Here,
The current $I = 5 A$

Time, $t = 30 s$

Voltage, $V = I \times R = 5 \times 20 = 100 V$

$\Rightarrow H = 100 \times 5 \times 30 = 1.5 \times 10^{4} J$

Therefore, the amount of heat developed in the electric iron is $1.5 \times 10^{4} J$ .

Question22

What determines the rate at which energy is delivered by a current?

Solution:

The rate at which energy is delivered by a current is called power. So, the power (of an appliance) determines the rate at which energy is delivered by a current.

Question23

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Solution:

Power (P) of the motor, $P = V I$ ,

where the voltage, $V = 220 V$ and the current, $I = 5 A$

Therefore, $P = 220 \times 5 = 1100 W$

Energy consumed by the motor $= P t$

Here, the time, $t = 2 hr = 2 \times 60 \times 60 = 7200 s$

Therefore, the energy $= 1100 \times 7200 = 7.92 \times 10^{6} J$

Questions

$= 1100 \times 7200 = 7.92 \times 10^{6} J$

Q1	What is a voltmeter?

Q2	What is the heating effect of current?

Q3	What is a power rating?

Q4	Define electric power?

Q5	A current of 0.5 A is passed through a filament of an electric bulb for 30 minutes. Find the amount of electric charge that flows through the circuit.

Q6	What is the purpose of an ammeter?

Q7	Define potential difference.

Q8	Explain the common application of Joule’s law of heating in the fuse used in electric circuits?

Q9	Describe how heating effect of current is used to produce light, as in an electric bulb.

Q10	Illustrate a schematic diagram of an electric circuit.

Q11	A current of 8 A from 250 V source, flows through an electric iron for two hours. Calculate the energy consumed in kWh.

Q12	Show a circuit diagram with resistors connected in parallel.

Q13	The potential difference between the terminals of an electric heater is 120 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 240 V?

Q14	A housewife uses a 200 W bulb for 4 hours a day and an electric heater of 600 W for 2 hours a day. What is the total cost of electrical energy consumption for the month of June at the rate of Rs. 4 per unit?

Q15	In the given figure, R₁= 20 Ω, R₂ = 80 Ω, R₃ = 60 Ω, R₄ = 40 Ω, R₅ = 120 Ω, and a 144 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Q16	In the circuit diagram, if the resistors R₁, R₂ and R₃ have the values 10 Ω, 20 Ω, 60 Ω, respectively, which have been connected to a battery of 24V, calculate: (a) The current through each resistor. (b) The total current in the circuit. (c) The total circuit resistance

Q17	Illustrate symbols of commonly used components in circuit diagrams?

Q18	Explain Ohm’s law.

Q19	What are the factors on which the resistance of a conductor depends?

Q20	Derive the expression for the combined resistance when the resistors are connected in a series.

Questions

Q1	What is a voltmeter?

A1	The potential difference is measured by means of an instrument called the voltmeter. The voltmeter is always connected in parallel across the points between which the potential difference is to be measured.

Q2	What is the heating effect of current?

A2	When an electric current passes through a high resistance wire, the wire becomes very hot. This phenomenon is known as heating effect of current.

Q3	What is a power rating?

A3	The power rating of an electrical appliance tells the rate at which electrical energy is consumed by the appliance. For example, an appliance with a power rating of 60 watts will consume energy at the rate of 60 joules per second.

Q4	Define electric power?

A4	The electric work done per unit time is called electric power. Power = work done/time taken. The SI unit of power is watt.

Q5	A current of 0.5 A is passed through a filament of an electric bulb for 30 minutes. Find the amount of electric charge that flows through the circuit.

A5	We are given, I = 0.5 A; t = 10 min = 1800 s. We have Q = It = 0.5 A × 1800 s = 900 C.

Q6	What is the purpose of an ammeter?

A6	An ammeter is an instrument used to measure electric current in a circuit. It is always connected in series in a circuit through which the current is to be measured.

Q7	Define potential difference.

A7	For flow of charges in a conducting metallic wire, the electrons move only if there is a difference of electric pressure called the potential difference along the conductor. This difference of potential may be produced by a battery, consisting of one or more electric cells.

Q8	Explain the common application of Joule’s law of heating in the fuse used in electric circuits?

A8

A fuse consists of a piece of wire made up of a metal or an alloy of appropriate melting point. It is placed in series with the device. It protects circuits and appliances by stopping the flow of excessive current through the circuit.

Whenever a current, which is larger than the specified value of the fuse, flows through the circuit, the temperature of the fuse wire increases. Due to low melting point, the fuse wire melts and the circuit breaks. Thus the excess current is not able to reach the appliance and it is thereby protected.

Q9	Describe how heating effect of current is used to produce light, as in an electric bulb.

A9

High resistance wire becomes very hot when electric current passes through it. This heating effect of electric current is used to glow an electric bulb.

The filament of an incandescent bulb is made of tungsten wire. Tungsten has a very high melting point. When current is allowed to pass through the filament, it becomes very hot but does not melt.

At around 2500^oC, the tungsten filament of an electric bulb becomes white hot and glows to emit light. Most of the power consumed by the filament is converted into heat, but a small part of it is radiated in the form of light.

Q10	Illustrate a schematic diagram of an electric circuit.

A10	A schematic diagram of an electric circuit comprises of cell, electric bulb, ammeter and plug key as are shown. When the key is inserted, the electric current flows in the circuit from the positive terminal of the cell to the negative terminal of the cell through the bulb and ammeter.

Q11	A current of 8 A from 250 V source, flows through an electric iron for two hours. Calculate the energy consumed in kWh.

A11	P = V I E= P x t E = VIt E = 250 x 8 x 2 = 4000 Wh = 4 kWh.

Q12	Show a circuit diagram with resistors connected in parallel.

A12	This figure shows a combination of resistors in which three resistors are connected together between the points A and B. Here, the resistors are said to be connected in parallel.

Q13	The potential difference between the terminals of an electric heater is 120 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 240 V?

A13	Potential difference, V = 120 V, current, I = 4 A. According to Ohm’s law, R= V/I = 120 or, V/4 A = 30 Ω. When the potential difference is increased to 240 V the current is given by, V/R = 240 or, V/30 Ω = 8 A The amount of current that the heater would draw is 8 A.

Q14	A housewife uses a 200 W bulb for 4 hours a day and an electric heater of 600 W for 2 hours a day. What is the total cost of electrical energy consumption for the month of June at the rate of Rs. 4 per unit?

A14	Number of units consumed in a month = [200/ 1000 x 4 + 600/1000 x 2]x 30 = (0.8 + 1.2) kWh x 30 = 2.0 kWh x 30 = 60 kWh Total cost = 60 x Rs. 4 = Rs. 240

Q15	In the given figure, R₁= 20 Ω, R₂ = 80 Ω, R₃ = 60 Ω, R₄ = 40 Ω, R₅ = 120 Ω, and a 144 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Q16	In the circuit diagram, if the resistors R₁, R₂ and R₃ have the values 10 Ω, 20 Ω, 60 Ω, respectively, which have been connected to a battery of 24V, calculate: (a) The current through each resistor. (b) The total current in the circuit. (c) The total circuit resistance

A16

R₁= 10 Ω, R₂ = 20 Ω, and R₃ = 60 Ω.

Potential difference across the battery, V = 24 V.

This is also the potential difference across each of the individual resistor.

To calculate the current in the resistors, we use Ohm’s law.

The current, I₁ through R₁ = V/ R₁

⇒I₁ = 24 V/10 Ω = 2.4 A.

The current I₂, through R₂ = V/ R₂

⇒I₂ = 24 V/20 Ω = 1.2 A.

The current I₃, through R₃ = V/R₃

⇒I₃ = 24 V/60 Ω = 0.4 A.

The total current in the circuit,

I = I₁+ I₂+ I₃ = (2.4 + 1.2 + 0.4) A

= 4 A

Let the total resistance be R_t.

(1/R_t) = (1/10) + (1/20) + (1/20)

= (1/6)

Thus, R_t = 6 Ω.

Q17	Illustrate symbols of commonly used components in circuit diagrams?

A17

Q18	Explain Ohm’s law.

A18

Georg Simon Ohm (1787–1854) found out the relationship between the current I, flowing through a metallic wire and the potential difference across its terminals. He found that the potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. This is called Ohm’s law. In other words, V = IR

A straight line plot shows that as the potential difference across the wire increases, the current flowing through the wire increases linearly.

The V–I graph is a straight line that passes through the origin of the graph, as shown in the figure. Thus, V/I is a constant ratio.

Q19	What are the factors on which the resistance of a conductor depends?

A19

Resistance of the conductor depends

(a) On its length,

(b) On its area of cross-section, and

The resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A). That is,

R = ρ(l/A)

where ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The metals and alloys are good conductors of electricity. The resistance of a material varies with change in temperature.

Q20	Derive the expression for the combined resistance when the resistors are connected in a series.

A20

Let’s consider a series as given. Let I be the current through the circuit.

The current through each resistor is same and is equal to I and the total voltage across the circuit, V =V₁+ V₂+ V₃

Considering three resistors joined in series by an equivalent single resistor of resistance R, the potential difference V across it, and the current I through the circuit would remain the same.

So we have V = IR (Ohm’s law)

On applying Ohm’s law to the three resistors separately, we get

V₁ = IR₁,V₂= IR₂ and V₃= IR₃

⇒ IR = IR₁+ IR₂+ IR₃

⇒ IR = I(R₁ + R₂ +R₃)

⇒ R = R₁+ R₂ + R_3.

ELECTRICITY CBSC CLASS 10TH F

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