SMART EDUCATIONS

5 SMART EDUCATIONS

Q. What is ‘R’ or Resistance of a wire?

Ans. Resistance of a wire is the property of a wire to resist the flow of current through it.

It is represented by the letter ‘R’. Its SI unit is . Ω( 𝑜ℎ )

𝑅 =

𝑉

𝐼

Q. What are the factors on which the resistance of a wire depends?

Ans. The resistance of a wire depends on the following four factors:-

1. Length of a wir e:- Resistance of a wire is directly proportional to the length of

the wire.

2. Area of the cross-section of a wire :- Resistance of a wire is inversely

proportional to the area of the cross-section of a wire.

3. Temperature of the wire :- Resistance of a wire is directly proportional to the

temperature of the wire.

4. Nature of the wire :- Resistance of a wire also depends on the nature of the wire

for example :- for silver it is lowest but for aluminum it is higher.

Q. What is the relation between the resistance of a wire, its length and its area of the

cross-section?

Ans. Because the resistance of a wire is directly proportional to the length of the wire.

𝑅 ∝ 𝑙 ( 𝑖 )

and it is inversely proportional to the area of the cross-section of a wire.

𝑅 ∝

1

𝐴

( 𝑖𝑖 )

From (i) and (ii), we get

𝑅 ∝

𝑙

𝐴

𝑅 =

ρ 𝑙

𝐴

( 𝑤ℎ𝑒𝑟𝑒 ρ 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑡𝑦 )

Q. What is the relation between ‘I’ and ‘R’?

Ans. ‘I’ is inversely proportional to ‘R’.

Q. What is (Rho) or What is resistivity?

' ρ '

Ans. Resistivity of a wire is the property of a wire to resist the flow of current through it.

But unlike the resistance, it does not depend on the length or cross-section of a wire.

Resistivity of a given wire at a given temperature is constant. It is represented by the

letter . Its SI unit is .

ρ Ω 𝑚

ρ=

𝑅𝐴

𝑙

SMART EDUCATIONS

6 SMART EDUCATIONS

Q. How much current will an electric bulb draw from a 220 V source, if the resistance of

the bulb filament is 1200 Ω? (b) How much current will an electric heater coil draw from

a 220 V source, if the resistance of the heater coil is 100 Ω?

Ans.

(a) (b)

𝐼 = ? 𝐼 = ?

𝑉 = 220 𝑉 𝑉 = 220 𝑉

𝑅 = 1200 Ω 𝑅 = 100 Ω

For(a) For(b)

𝑉 = 𝐼𝑅 𝑉 = 𝐼𝑅

220 = 𝐼 × 1200 220 = 𝐼 × 100

𝐼 =

220

1200

𝐼 =

220

100

𝐼 = 0 . 18 𝐴 𝐼 = 2 . 2 𝐴

Q. The potential difference between the terminals of an electric heater is 60 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 120 V?

Ans.

In the 1

st

case In the 2

nd

case

𝑉 = 60 𝑉 𝑉 = 120 𝑉

𝐼 = 4 𝐴 𝐼 = ?

For the 1

st

case

𝑉 = 𝐼𝑅

60 = 4 × 𝑅

𝑅 =

60

4

𝑅 = 15 Ω

For the 2

nd

case

Because the same heater was used in the 2

nd

case so, the value of R will be be same

as of 1

st

case i.e

𝑅 = 15 Ω

𝑉 = 𝐼𝑅

120 = 𝐼 × 15

𝐼 =

120

15

𝐼 = 8 𝐴

Q. Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is

0.3 mm, what will be the resistivity of the metal at that temperature?

SMART EDUCATIONS

7 SMART EDUCATIONS

Ans.

𝑙 = 1 𝑚

𝑅 = 26 Ω

𝑑 = 0 . 3 𝑚𝑚

𝑟 =

0 . 3

2

𝑚𝑚

= 0 . 15 × 10

− 3

𝑚

ρ= ?

ρ=

𝑅𝐴

𝑙

ρ =

20 × π 𝑟

2

1

ρ= 26 × 3 . 14 × 0 . 15 × 10

− 3

( )

2

ρ= 81 . 64 × 0 . 15 × 0 . 15 × 10

− 3

× 10

− 3

ρ= 1 . 836 × 10

− 6

ρ= 1 . 84 × 10

− 6

Ω 𝑚

Q. A wire of given material having length l and area of cross-section A has a resistance

of 4 Ω. What would be the resistance of another wire of the same material having length

l/2 and area of cross-section 2A?

Ans.

In the 1

st

case In the 2

2d

case

𝑙 = 𝑙 𝑙 =

𝑙

2

𝐴 = 𝐴 𝐴 = 2 𝐴

𝑅 = 4 Ω 𝑅 = ?

For the 1

st

case

ρ=

𝑅𝐴

𝑙

ρ=

4 × 𝐴

𝑙

ρ=

4 𝐴

𝑙

Because the same wire was used in 2

nd

case so the value of will be the same in the

ρ

2

nd

case.

For the 2

2d

case

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8 SMART EDUCATIONS

𝑅 =

ρ 𝑙

𝐴

𝑅 =

4 𝐴

𝑙

×

𝑙

2

2 𝐴

𝑅 =

2 𝐴

𝑅 = 1 Ω

Q. Will current flow more easily through a thick wire or a thin wire of the same material,

when connected to the same source? Why?

Ans. If a thick wire and a thin wire of the same material is connected to the same

source then through the thick wire more current will flow because the current is directly

proportional to the area of the cross-section of a wire and the thick wire has the greater

area of the cross-section.

Q. Let the resistance of an electrical component remain 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?

Ans. According to Ohm's law, current is directly proportional to the potential difference

hence if the potential difference across the two ends of the component decreases to half

of its former value then the current will also decrease to half of its former value.

Q. Why are coils of electric toasters and electric irons made of an alloy rather than a

pure metal?

Ans. Coils of electric toasters and electric irons are made of an alloy rather than a

pure metal because an alloy has high resistivity because of which it produces high

heating effect and moreover it does not melt easily even at high temperature.

Q. What are the ways to connect an electronic component to a circuit?

Ans. There are two ways to connect an electronic component to a circuit:-

1. Series Connection

2. Parallel Connection

Q. What does Series Connection mean?

Ans. Series Connection is a type of connection in which the second end of the first

resistor is connected to the first end of the second resistor and the second end of the

second resistor is connected to the first end of the third resistors and so on. This type of

connection is also known as end to end connection.

SMART EDUCATIONS

9 SMART EDUCATIONS

Q. What does Parallel Connection mean?

Ans. Parallel Connection is a type of connection in which the first end of the all resistors

are connected together and the second end of the all resistors are connected together.

Q. Derive the equivalent resistance in series combination.

Ans. Let the three resistors R

1

, R

2

, and R

3

are connected in series combination in a

circuit with 2 cells, a key, an ammeter in series, three voltmeters in parallel across the

three resistors and one voltmeter in parallel across the two cells as shown below:-

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝐴 = 𝐼

( 𝐼𝑛 𝑠𝑒𝑟𝑖𝑒𝑠 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 , ' 𝐼 ' 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 .)

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝑉 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑐𝑒𝑙𝑙 = 𝑉

𝑡

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝑉 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑅

1

= 𝑉

1

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝑉 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑅

2

= 𝑉

2

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝑉 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑅

3

= 𝑉

3

𝐻𝑒𝑛𝑐𝑒 𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑜ℎ𝑚 ' 𝑠 𝑙𝑎𝑤 :−

𝑉

𝑡

= 𝐼 𝑅

𝑡

𝑉

1

= 𝐼 𝑅

1

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10 SMART EDUCATIONS

𝑉

2

= 𝐼 𝑅

2

𝑉

3

= 𝐼 𝑅

3

𝑉

1

+ 𝑉

2

+ 𝑉

3

= 𝑉

𝑡

𝐼 𝑅

1

+ 𝐼 𝑅

2

+ 𝐼 𝑅

3

= 𝐼 𝑅

𝑡

𝐼 𝑅

1

+ 𝑅

2

+ 𝑅

3

( )

= 𝐼 𝑅

𝑡

𝑅

1

+ 𝑅

2

+ 𝑅

3

( )

=

𝐼 𝑅

𝑡

𝐼

𝑅

𝑡

= 𝑅

1

+ 𝑅

2

+ 𝑅

3

Hence the equivalent resistance in series combination is

𝑅

𝑡

= 𝑅

1

+ 𝑅

2

+ 𝑅

3

.......

Q. Derive the equivalent resistance in parallel combination.

Ans. Let the three resistors R

1

, R

2

, and R

3

are connected in parallel combination in a

circuit with 2 cells, a key, an ammeter in series, three another ammeter in series with the

three resistors and one voltmeter in parallel across the three resistors as shown below:-

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝑉 = 𝑉

( 𝐼𝑛 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 , ' 𝑉 ' 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 .)

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝐴 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 = 𝐼

𝑡

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝐴 𝑖𝑛 𝑡ℎ𝑒 𝑅

1

= 𝐼

1

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝐴 𝑖𝑛 𝑡ℎ𝑒 𝑅

2

= 𝐼

2

𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑜𝑓 𝐴 𝑖𝑛 𝑡ℎ𝑒 𝑅

3

= 𝐼

3

𝐻𝑒𝑛𝑐𝑒 𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑜ℎ𝑚 ' 𝑠 𝑙𝑎𝑤 :−

𝐼

𝑡

=

𝑉

𝑅

𝑡

𝐼

1

=

𝑉

𝑅

1

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11 SMART EDUCATIONS

𝐼

2

=

𝑉

𝑅

2

𝐼

3

=

𝑉

𝑅

3

𝐼

𝑡

= 𝐼

1

+ 𝐼

2

+ 𝐼

3

𝑉

𝑅

𝑡

=

𝑉

𝑅

1

+

𝑉

𝑅

2

+

𝑉

𝑅

3

𝑉

𝑅

𝑡

= 𝑉

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

( )

1

𝑅

𝑡

=

𝑉

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

( )

1

𝑅

𝑡

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

Hence the equivalent resistance in parallel combination is :-

1

𝑅

𝑡

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

...........

Q. What are the differences between a series combination and a parallel combination?

Ans.

Series combination

Parallel combination

The second end of the first resistor is

connected to the first end of the second

resistor and the second end of the second

resistor is connected to the first end of the

third resistors and so on.

The first end of the all resistors are

connected together and the second end

of the all resistors are connected together.

The total resistance is more than the

highest resistance in the circuit.

The total resistance is less than the

smallest resistance in the circuit.

If there is fault in any of the connected

devices, the circuit will break and no other

devices will work.

If there is fault in any of the connected

devices, the circuit will not break and no other

devices will stop work other than the fault

component.

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12 SMART EDUCATIONS

It increases the temperature of the circuit.

It decreases the temperature of the circuit.

All devices take equal current.

All devices take current accordingly.

All the devices are connected with one switch

only, so can't be turned off any one of the

devices manually.

All the devices are connected with different

switches, so any one of the devices can be

turned off manually.

Q. How does a series combination and a parallel combination affect the total resistance

in a circuit?

Ans. Series combination increases the total resistance while parallel combination

decreases the total resistance. In a series combination the total resistance is more than

the highest resistance in the circuit and in a parallel combination the total resistance is

less than the smallest resistance in the circuit.

Q. An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are

connected to a 6 V battery . Calculate (a) the total resistance of the circuit, (b) the

current through the circuit, and (c) the potential difference across the electric lamp and

conductor.

Ans.

𝑅

1

= 20 Ω ( 𝑙𝑎𝑚𝑝 )

𝑅

2

= 4 Ω ( 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 )

𝑉 = 6 𝑉

𝑅

𝑡

= ?

𝐼

𝑡

= ?

𝑉

1

= ?

Because this is a series combination, hence

𝑅

𝑡

= 𝑅

1

+ 𝑅

2

𝑅

𝑡

= 20 + 4

Ans(a)

𝑅

𝑡

= 24 Ω

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13 SMART EDUCATIONS

𝐼

𝑡

=

𝑉

𝑅

𝑡

𝐼

𝑡

=

6

24

Ans(b)

𝐼

𝑡

= 0 . 25 𝐴

𝑉

1

= 𝐼 𝑅

1

𝑉

1

= 0 . 25 × 20

Ans(c)

𝑉

1

= 5 𝑉

Q. 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. Put 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?

Ans.

Because this is a series combination, hence:-

𝑅

𝑡

= 𝑅

1

+ 𝑅

2

+ 𝑅

3

𝑅

𝑡

= 5 + 8 + 12

𝑅

𝑡

= 25 Ω

𝐼

𝑡

=

𝑉

𝑅

𝑡

𝐼

𝑡

=

2 + 2 + 2

25

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14 SMART EDUCATIONS

𝐼

𝑡

= 0 . 24 𝐴

Hence the reading in the ammeter = 0.24 A Ans.

𝑉

3

= 𝐼 𝑅

3

𝑉

3

= 0 . 24 × 12

𝑉

3

= 2 . 88 𝑉

Hence the reading in the voltmeter = 0.24 A Ans.

Q. In the given circuit, suppose the resistors R

1

, R

2

and R

3

have the values 5 Ω, 10 Ω,

30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the

current through each resistor, (b) the total current in the circuit, and (c) the total circuit

resistance.

Ans.

𝑅

1

= 5 Ω

𝑅

2

= 10 Ω

𝑅

3

= 30 Ω

𝑉 = 12 𝑉

(a)

𝐼

1

= ? 𝐼

2

= ? 𝐼

3

= ?

(b)

𝐼

𝑡

= ?

(c)

𝑅

𝑡

= ?

𝐼

1

=

𝑉

𝑅

1

𝐼

1

=

𝑉

𝑅

1

𝐼

1

=

𝑉

𝑅

1

𝐼

1

=

12

5

𝐼

1

=

12

10

𝐼

1

=

12

30

𝐼

1

= 2 . 4 𝐴 𝐼

1

= 1 . 2 𝐴 𝐼

1

= 0 . 4 𝐴

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15 SMART EDUCATIONS

Ans(a)

𝐼

𝑡

= 𝐼

1

+ 𝐼

2

+ 𝐼

3

𝐼

𝑡

= 2 . 4 + 1 . 2 + 0 . 4

Ans(b)

𝐼

𝑡

= 4 𝐴

For R

t

we can use two methods:-

Method

1

:-

Because this is parallel combination, hence

1

𝑅

𝑡

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

1

𝑅

𝑡

=

1

5

+

1

10

+

1

30

1

𝑅

𝑡

=

6 + 3 + 1

30

1

𝑅

𝑡

=

10

30

10 𝑅

𝑡

= 30

𝑅

𝑡

=

30

10

Ans(c)

𝑅

𝑡

= 3 Ω

Method

2

:-

𝑅

𝑡

=

𝑉

𝐼

𝑡

𝑅

𝑡

=

12

4

Ans(c)

𝑅

𝑡

= 3 Ω

Q. If in the given Fig, R

1

= 10 Ω, R

2

= 40 Ω, R

3

= 30 Ω, R

4

= 20 Ω, R

5

= 60 Ω, and a 12 V

battery is connected to the arrangement. Calculate (a) the total resistance in the circuit,

and (b) the total current flowing in the circuit.

SMART EDUCATIONS

16 SMART EDUCATIONS

Ans. R

1

= 10 Ω, R

2

= 40 Ω, R

3

= 30 Ω, R

4

= 20 Ω, R

5

= 60 Ω,

V = 12 V

(a) (b)

𝑅

𝑇

= ? 𝐼

𝑇

= ?

From the figure it is clear that R

1

and R

2

are connected in parallel and R

3

, R

4

and R

5

are

connected in another parallel connection. Let the total resistance for R

1

and R

2

is R

T1

and the total resistance for R

3

, R

4

and R

5

is R

T2

. So R

T1

and R

T2

are connected in series.

1

𝑅

𝑇 1

=

1

𝑅

1

+

1

𝑅

2

1

𝑅

𝑇 2

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

1

𝑅

𝑇 1

=

1

10

+

1

40

1

𝑅

𝑇 2

=

1

30

+

1

20

+

1

60

1

𝑅

𝑇 1

=

4 + 1

40

1

𝑅

𝑇 2

=

2 + 3 + 1

60

1

𝑅

𝑇 1

=

5

40

1

𝑅

𝑇 2

=

6

60

𝑅

𝑇 1

= 8 Ω 𝑅

𝑇 2

= 10 Ω

𝑅

𝑇

= 𝑅

𝑇 1

+ 𝑅

𝑇 2

𝑅

𝑇

= 8 + 10

Ans(a)

𝑅

𝑇

= 18 Ω

𝐼

𝑇

=

𝑉

𝑅

𝑇

𝐼

𝑇

=

12

18

Ans(b)

𝐼

𝑇

= 0 . 67 𝐴

Q. 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

SMART EDUCATIONS

17 SMART EDUCATIONS

Ans. As we know that in a parallel combination the total resistance is less than the

smallest resistance in the circuit so the equivalent resistance in (a) is less than 1Ω and

in (b) is less than 1 Ω

For (a)

R

1

= 1 Ω and R

2

= 10

6

Ω

For parallel connection:-

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

1

𝑅

𝑇

=

1

+

1

10

6

1

𝑅

𝑇

=

10

6

+ 1

10

6

𝑅

𝑇

=

10

6

10

6

+ 1

As denominator of more than the numerator hence

10

6

10

6

+ 1

10

6

10

6

+ 1

< 1

𝑅

𝑇

< 1

For (b)

R

1

= 1 Ω, R

2

= 10

3

Ω and R

2

= 10

6

Ω

For parallel connection:-

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

1

𝑅

𝑇

=

1

+

1

10

3

+

1

10

6

1

𝑅

𝑇

=

10

6

+ 10

3

+ 1

10

6

𝑅

𝑇

=

10

6

10

6

+ 10

3

+ 1

As denominator of more than the numerator hence

10

6

10

6

+ 10

3

+ 1

10

6

10

6

+ 10

3

+ 1

< 1

𝑅

𝑇

< 1

SMART EDUCATIONS

18 SMART EDUCATIONS

Q. 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?

Ans.

R

1

= 100 Ω, R

2

= 50 Ω and R

3

= 500 Ω

V = 220 V

R

T

= ?

I

T

= ?

For parallel connection:-

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

1

𝑅

𝑇

=

1

100

+

1

50

+

1

500

1

𝑅

𝑇

=

5 + 10 + 1

500

1

𝑅

𝑇

=

16

500

𝑅

𝑇

=

500

16

Ans(i)

𝑅

𝑇

= 31 . 25 Ω

𝐼

𝑇

=

𝑉

𝑅

𝑇

𝐼

𝑇

=

220

31 . 25

Ans(ii)

𝐼

𝑇

= 7 . 04 𝐴

Q. What are the advantages of connecting electrical devices in parallel with the battery

instead of connecting them in series?

Ans. There are the following advantages of connecting electrical devices in parallel :-

1. All devices take current accordingly.

2. Separate switches can be applied for every single device.

3. Total resistance decreases and the current moves freely.

4. Less energy is wasted in terms of heat.

SMART EDUCATIONS

19 SMART EDUCATIONS

5. If there is fault in any of the connected devices, the circuit will not break and no other

devices will stop work other than the fault component.

Q. 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?

Ans. (a) 4 Ω

If we connect all three resistances in series then the total resistance will be more than 6

Ω and if we connect all three resistances in parallel then the total resistance will be less

than 2 Ω, it means to get 4 Ω we must combine the two combinations.

Let say, 3 Ω, and 6 Ω are connected in parallel and the equivalent resistance of

the two is connected in series with 2 Ω.

For 3Ω and 6Ω in parallel

1

𝑅

𝑇 1

=

1

𝑅

1

+

1

𝑅

2

1

𝑅

𝑇 1

=

1

3

+

1

6

1

𝑅

𝑇 1

=

2 + 1

6

1

𝑅

𝑇 1

=

3

6

𝑅

𝑇 1

= 2 Ω

Now the equivalent of 3 Ω and 6 Ω with 2 Ω in series:-

𝑅

𝑇

= 𝑅

𝑇 1

+ 𝑅

3

𝑅

𝑇

= 2 + 2

Ans(a)

𝑅

𝑇

= 4 Ω

(b) 1 Ω

If we connect all the three resistances in parallel then the total resistance will be

less than 2 Ω, it means may be 1 Ω

SMART EDUCATIONS

20 SMART EDUCATIONS

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

1

𝑅

𝑇

=

1

2

+

1

3

+

1

6

1

𝑅

𝑇

=

3 + 2 + 1

6

1

𝑅

𝑇

=

6

𝑅

𝑇

= 1

Ans(b)

𝑅

𝑇

= 1 Ω

Q. 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?

Ans. (a) To get the highest resistance all the resistance must be connected in

series:-

𝑅

𝑇

= 𝑅

1

+ 𝑅

2

+ 𝑅

3

+ 𝑅

4

𝑅

𝑇

= 4 + 8 + 12 + 24

Ans(a)

𝑅

𝑇

= 48 Ω

(b) To get the lowest resistance all the resistance must be connected in parallel:-

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

+

1

𝑅

4

1

𝑅

𝑇

=

1

4

+

1

8

+

1

12

+

1

24

1

𝑅

𝑇

=

6 + 3 + 2 + 1

24

1

𝑅

𝑇

=

12

24

𝑅

𝑇

= 2

Ans(b)

𝑅

𝑇

= 2 Ω

Q. What does the heating effect of electric current mean?

SMART EDUCATIONS

21 SMART EDUCATIONS

Ans. When an electrical device is connected to a source of electricity then the device

uses the energy of the source to work and some or large amount of the energy is used

by the device to heat itself. So the heat produced by the electric current is called the

heating effect of electric current. The amount of heat produced is denoted as ‘H’. The SI

unit of heat is Joule(J).

𝐻 = 𝑉𝐼𝑇

Q. What are the factors on which the heating effect of electric current depends?

Ans. The heating effect of electric current depends on the following factors:-

1. Resistance of the device.

2. Power of the device.

3. Current in the device.

4. Potential difference in the device.

5. Time at which the device is operated.

Q. What is joule's law of heating?

Ans. Joule’s law of heating states that:-

1. The heat produced in a resistor is directly proportional to the square of the

electric current it takes.

2. The heat produced in a resistor is directly proportional to the resistance of the

resistor.

3. The heat produced in a resistor is directly proportional to the time at which the

device is operated.

The expression for Joule’s law of heating is as follows

𝐻 = 𝐼

2

𝑅𝑇

Q. Write the applications of the heating effect of electric current.

Ans. Some amount of electric energy is used in terms of heat by the device but

according to Joule’s law of heating, if we use a highly resistive device then the whole

amount of the energy will be used in terms of heat and the produced heat can be used

for many purposes such as:-.

1. Used in electric iron.

2. Used in electric toasters.

3. Used in electric heaters.

4. Used in electric ovens.

5. Used in electric kettles.

6. Used in electric bulbs.

7. Used in electric fuses.

Q . Write all the expressions for heat.

SMART EDUCATIONS

22 SMART EDUCATIONS

Ans.

𝐻 = 𝑉𝐼𝑇

𝐻 = 𝐼

2

𝑅𝑇

𝐻 = 𝑃𝑇

𝐻 = 𝑊

𝐻 = 𝑉𝑄

Q. How is the heating effect of electric current used for lighting?

Ans. If electric current is passed through a conductor then it becomes hot due to the

heating effect of electric current but if the conductor is highly resistive then it produces

so much heat that it starts glowing and produces light. Same thing happens in the case

of an electric bulb. In the case of an electric bulb the highly resistive conductor is

tungsten which has not only high resistance but has high melting which prevents it from

melting at such high temperature. In addition to this the bulb is filled with argon gas

which helps the tungsten to not to corrode and hence prolong its life.

Q. What is an electric fuse?

Ans. An electric fuse is a safety device for the circuit and its components. It works on

the principle of Joule’s law of heating. It is connected in series. The fuses used for

domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc.

Q. How does an electric fuse work?

Ans. The fuse is placed in series with the device. It consists of a piece of wire made of

a metal or an alloy of appropriate melting point, for example aluminum, copper, iron,

lead etc. If a current larger than the specified value flows through the circuit, the

temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.

Q. How to calculate which rating of fuse must be installed in a circuit?

Ans. To calculate the rating of a fuse to which must be installed, first of all find the sum

of power of the devices which will be connected in the circuit and the potential

difference of the source and then when we divide the sum to the potential difference

then it gives the rating of fuse.

For example, For an electric iron which consumes 1 kW electric power when

operated at 220 V, a current of (1000/220) A, that is, 4.54 A will flow in the circuit. In this

case, a 5 A fuse must be used.

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

Ans. Because the heating element is highly resistive so it produces so much heat that it

starts glowing but the cord of an electric heater is not so resistive that it may produce so

much heat.

SMART EDUCATIONS

23 SMART EDUCATIONS

Q. Compute the heat generated while transferring 96000 coulomb of charge in one hour

through a potential difference of 50 V.

Ans.

𝑄 = 96000 𝐶

𝑡 = 1 ℎ

= 1 × 60 × 60 𝑠

= 3600 𝑠

𝑉 = 50 𝑉

Ans. 𝐻 = 𝑉𝑄

𝐻 = 50 × 96000

𝐻 = 4800000 𝐽

Or

𝐻 = 4800 𝐾𝐽

Ans.

𝐻 = 4 . 8 × 10

6

𝐽

Q. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat

developed in 30 s.

Ans.

𝑅 = 20 Ω

𝐼 = 5 𝐴

𝑡 = 30 𝑠

𝐻 = ?

𝐻 = 𝐼

2

𝑅𝑇

𝐻 = 5

2

× 20 × 30

𝐻 = 15000 𝐽

Ans.

𝐻 = 15 𝐾𝐽

Q. What is electric Power?

Ans. The rate of consuming electric energy by a device is known as the power of the

device. The electric power is denoted as ‘P’. The SI unit of electric power is watt(W).

𝑃 =

𝑊

𝑇

𝑃 = 𝑉𝐼

Q. Define 1 watt.

Ans. 1 watt is the power of an electric device which uses 1 joule of energy per second.

1 𝑤𝑎𝑡𝑡 =

1 𝑗𝑜𝑢𝑙𝑒

1 𝑠𝑒𝑐𝑜𝑛𝑑

OR

SMART EDUCATIONS

24 SMART EDUCATIONS

1 watt is the power of an electric device which takes 1A of current when operated at 1V.

1 𝑤𝑎𝑡𝑡 = 1 𝐴 × 1 𝑉

Q. What is the unit of electrical energy?

Ans. Unit for all types of energy is Joule, so the unit of electrical energy is also Joule.

But for commercial use, the unit of electrical energy is kilowatt hour, which is also known

as ‘unit’.

Q. What does 1 unit mean?

Ans. 1 unit = 1 Kwh, or

1 𝑢𝑛𝑖𝑡 = 1 𝐾𝑤ℎ

= 1 × 1000 × 3600 𝐽

1 𝑢𝑛𝑖𝑡 = 3 . 6 × 10

6

𝐽

Q. An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the

power of the bulb?

Ans.

𝑉 = 220 𝑉

𝐼 = 0 . 50 𝐴

𝑃 = ?

𝑃 = 𝑉𝐼

𝑃 = 220 × 0 . 50

𝑃 = 110 𝑊

Q. An electric refrigerator rated 400 W operates 8 hours/day. What is the cost of the

energy to operate it for 30 days at Rs 3.00 per Kwh?

Ans.

𝑃 = 400 𝑊

= 0 . 4 𝐾𝑤

𝑡 = 8 ℎ / 𝑑𝑎𝑦

𝑑𝑎𝑦𝑠 = 30 𝑑𝑎𝑦𝑠

𝑟𝑎𝑡𝑒 = 3 𝑝𝑒𝑟 𝐾𝑤ℎ

𝑐𝑜𝑠𝑡 = ?

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐𝑜𝑠𝑡 = 𝑃 × 𝑡 × 𝑑𝑎𝑦𝑠 × 𝑟𝑎𝑡𝑒

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐𝑜𝑠𝑡 = 0 . 4 × 8 × 30 × 3

SMART EDUCATIONS

25 SMART EDUCATIONS

Ans.

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐𝑜𝑠𝑡 = 288 𝑅𝑠

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

Ans. Electric Power.

Q. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and

the energy consumed in 2 h.

Ans.

𝐼 = 5 𝐴

𝑉 = 220 𝑉

𝑡 = 2 ℎ

𝑃 = ?

𝐻 = ?

𝑃 = 𝑉𝐼

𝑃 = 220 × 5

Ans 𝑃 = 1100 𝑊

𝑃 = 1 . 1 𝐾𝑤

𝐻 = 𝑃𝑡

𝐻 = 1 . 1 × 2

𝐻 = 2 . 2 𝐾𝑤ℎ

𝐻 = 2 . 2 × 3 . 6 × 10

6

𝐽

Ans

𝐻 = 7 . 92 × 10

6

𝐽

EXERCISES

Q. 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

is –

𝑅

𝑅 '

Ans. When a wire of resistance R is cut into five equal parts then resistance of each

part is because resistance of a wire is directly proportional to the length of a wire.

𝑅

5

SMART EDUCATIONS

26 SMART EDUCATIONS

When 5 resistors of are connected in parallel then:-

𝑅

5

Ω

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

+

1

𝑅

4

+

1

𝑅

5

1

𝑅 '

=

1

𝑅

5

+

1

𝑅

5

+

1

𝑅

5

+

1

𝑅

5

+

1

𝑅

5

1

𝑅 '

=

5

𝑅

+

5

𝑅

+

5

𝑅

+

5

𝑅

+

5

𝑅

1

𝑅 '

=

5 + 5 + 5 + 5 + 5

𝑅

1

𝑅 '

=

25

𝑅

𝑅 ' =

𝑅

25

Ω

So,

𝑅

𝑅 '

=

𝑅

25

𝑅

𝑅 '

=

𝑅 × 25

𝑅

Ans

𝑅

𝑅 '

=

25

1

Q. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power

consumed will be –

Ans.

𝑉

1

= 220 𝑉

𝑃

1

= 100 𝑊

But if

𝑉

2

= 110 𝑉

𝑃

2

= ?

For the first case:-

𝑃 = 𝑉𝐼

SMART EDUCATIONS

27 SMART EDUCATIONS

100 = 220 × 𝐼

𝐼 =

100

220

𝐼 =

5

11

𝐴

For the second case:-

𝑉

2

= 110 𝑉 ( 𝐼𝑡 𝑖𝑠 𝑐𝑙𝑒𝑎𝑟𝑙𝑦 ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 )

𝐼

2

=

1

2

𝑜𝑓 𝐼

1

( 𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑜ℎ𝑚 ' 𝑠 𝑙𝑎𝑤 )

𝐼

2

=

1

2

×

5

11

𝐼

2

=

5

22

Hence:-

𝑃 = 𝑉𝐼

𝑃 = 110 ×

5

22

Ans

𝑃 = 25 𝑊

OR 2

nd

Method

For the first case:-

𝑃 =

𝑉

2

𝑅

100 =

220

2

𝑅

100 𝑅 = 48400

𝑅 =

48400

100

𝑅 = 484 Ω

For the second case:-

Because the same bulb is to be used, hence the resistance of the bulb will be same as

that of 484

Ω.

𝑃 =

𝑉

2

𝑅

𝑃 =

110

2

484

𝑃 =

12100

484

Ans.

𝑃 = 25 𝑊

SMART EDUCATIONS

28 SMART EDUCATIONS

Q. 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 –

Ans.

As the two wires are of the same materials, equal length and equal diameters,

hence the resistance of the two wires will be the same.

R

1

= R

R

2

= R

V = V

𝐻

𝑠

𝐻

𝑃

= ?

For series connection:-

𝑅

𝑇

= 𝑅

1

+ 𝑅

2

𝑅

𝑇

= 𝑅 + 𝑅

𝑅

𝑇

= 2 𝑅

𝐻

𝑆

=

𝑉

2

𝑡

𝑅

𝐻

𝑆

=

𝑉

2

𝑡

2 𝑅

For parallel connection:-

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

1

𝑅

𝑇

=

1

𝑅

+

1

𝑅

1

𝑅

𝑇

=

1 + 1

𝑅

1

𝑅

𝑇

=

2

𝑅

𝑇

=

𝑅

2

𝐻

𝑃

=

𝑉

2

𝑡

𝑅

𝐻

𝑃

=

𝑉

2

𝑡

𝑅

2

SMART EDUCATIONS

29 SMART EDUCATIONS

𝐻

𝑃

=

2 𝑉

2

𝑡

𝑅

Hence,

𝐻

𝑠

𝐻

𝑃

=

𝑉

2

𝑡

2 𝑅

2 𝑉

2

𝑡

𝑅

𝐻

𝑠

𝐻

𝑃

=

𝑉

2

𝑡

2 𝑅

×

𝑅

2 𝑉

2

𝑡

𝐻

𝑠

𝐻

𝑃

=

1

4

H

s

: H

P

= 1 : 4 Ans.

Q. A copper wire has a diameter 0.5 mm and resistivity of 1.6 × 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?

Ans.

𝑑 = 0 . 5 𝑚𝑚

𝑟 = 0 . 25 𝑚𝑚

= 0 . 25 × 10

− 3

𝑚

ρ= 1 . 6 × 10

− 8

Ω 𝑚

𝑙 = ?

𝑅 = 10 Ω

But if,

𝑑 = 0 . 5 × 2 𝑚𝑚

𝑅 = ?

𝑅 =

ρ 𝑙

𝐴

10 =

1 . 6 × 10

− 8

× 𝑙

π 𝑟

2

10 =

1 . 6 × 10

− 8

× 𝑙

3 . 14 × 0 . 25 × 10

− 3

( )

2

10 =

1 . 6 × 10

− 8

× 𝑙

3 . 14 × 0 . 0625 × 10

− 6

10 =

1 . 6 × 10

− 8 + 6

× 𝑙

3 . 14 × 0 . 0625

SMART EDUCATIONS

30 SMART EDUCATIONS

1 . 6 × 10

− 2

× 𝑙 = 10 × 3 . 14 × 0 . 0625

𝑙 =

31 . 4 × 0 . 0625

1 . 6 × 10

− 2

𝑙 =

31 . 4 × 0 . 0625 × 10

2

1 . 6

𝑙 =

31 . 4 × 0 . 0625 × 100

1 . 6

𝑙 =

196 . 25

1 . 6

Ans(i)

𝑙 = 122 . 66 𝑚

But now the diameter has been doubled, hence the radius will also be doubled. Then:-

𝑅 ' =

ρ 𝑙

𝐴 '

𝑅 ' =

ρ × 𝑙

π 𝑟 '

2

𝑅 ' =

ρ × 𝑙

π × 2 𝑟 ( )

2

𝑅 ' =

ρ × 𝑙

π × 4 𝑟

2

𝑅 ' =

1

4

×

ρ × 𝑙

π 𝑟

2

𝑅 ' =

1

4

× 𝑅

𝑅 ' =

1

4

× 10

Ans(ii)

𝑅 ' = 2 . 5 Ω

Q. 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.

Ans.

SMART EDUCATIONS

31 SMART EDUCATIONS

𝑅

1

=

𝑉

1

𝐼

1

𝑅

2

=

𝑉

2

𝐼

2

𝑅

3

=

𝑉

3

𝐼

3

𝑅

4

=

𝑉

4

𝐼

4

𝑅

5

=

𝑉

5

𝐼

5

𝑅

1

=

1 . 6

0 . 5

𝑅

2

=

3 . 4

1 . 0

𝑅

3

=

6 . 7

2 . 0

𝑅

4

=

10 . 2

3 . 0

𝑅

5

=

13 . 2

4

𝑅

1

= 3 . 2 𝑅

2

= 3 . 4 𝑅

3

= 3 . 35 𝑅

4

= 3 . 4 𝑅

5

= 3 . 3

So,

𝑅

𝑎𝑣

=

3 . 2 + 3 . 4 + 3 . 35 + 3 . 4 + 3 . 3

5

Ans

𝑅

𝑎𝑣

= 3 . 33 Ω

Q. 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.

Ans.

𝑉 = 12 𝑉

𝐼 = 2 . 5 𝑚𝐴

= 2 . 5 × 10

− 3

𝐴

𝑅 = ?

𝑅 =

𝑉

𝐼

𝑅 =

12

2 . 5 × 10

− 3

𝑅 =

12 × 10

3

2 . 5

SMART EDUCATIONS

32 SMART EDUCATIONS

𝑅 =

12000

2 . 5

Ans 𝑅 = 4800 Ω

Q. 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 resistor?

Ans.

𝑉 = 9 𝑉

𝑅

1

= 0 . 2 Ω

𝑅

2

= 0 . 3 Ω

𝑅

3

= 0 . 4 Ω

𝑅

4

= 0 . 5 Ω

𝑅

5

= 12 Ω

𝐼 = ?

Because they are connected in series, hence:-

𝑅

𝑇

= 𝑅

1

+ 𝑅

2

+ 𝑅

3

+ 𝑅

4

+ 𝑅

5

𝑅

𝑇

= 0 . 2 + 0 . 3 + 0 . 4 + 0 . 5 + 12

𝑅

𝑇

= 13 . 4 Ω

𝐼 =

𝑉

𝑅

𝐼 =

9

13 . 4

Ans

𝐼 = 0 . 67 𝐴

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

Ans.

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟𝑠 = 𝑥

𝑅

1

= 𝑅

2

= 𝑅

3

...... 𝑅

𝑥

= 176 Ω

𝐼 = 5 𝐴

𝑉 = 220 𝑉

𝑅

𝑇

=

𝑉

𝐼

𝑅

𝑇

=

220

5

SMART EDUCATIONS

33 SMART EDUCATIONS

𝑅

𝑇

= 44 Ω

Because they are in parallel, hence:-

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

+

1

𝑅

3

....... +

1

𝑅

𝑥

1

44

=

1

176

+

1

176

+

1

176

....... + 𝑥 𝑡𝑖𝑚𝑒𝑠

1

44

=

1

176

× 𝑥

𝑥 =

176

44

𝑥 = 4

Hence total 4 resistors of each 176 are required to carry 5 A on a 220 V line.

Ω

Q. 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.

Ans.

R

1

= 6Ω, R

2

= 6 Ω and R

3

= 6 Ω

(i) R

T

= 9 Ω (How ?)

(ii) R

T

= 4 Ω (How ?)

(i) Let R

1

and R

2

are connected in parallel and the equivalent of the two resistors is

connected to the third in series as shown in figure:-

Equivalent resistance of the R

1

and R

2

:-

1

𝑅

𝑡

=

1

𝑅

1

+

1

𝑅

2

1

𝑅

𝑡

=

1

6

1

+

1

6

1

𝑅

𝑡

=

1 + 1

6

SMART EDUCATIONS

34 SMART EDUCATIONS

1

𝑅

𝑡

=

2

6

1

R

t

= 3

Ω

Now R

t

is connected with R

3

in series:-

R

T

= R

t

+ R

3

R

T

= 3 + 6

R

T

= 9 Ω Ans(i)

(i) Let R

1

and R

2

are connected in series and the equivalent of the two resistors is

connected to the third in parallel as shown in figure:-

For series combination:-

R

t

= R

1

+ R

2

R

t

= 6 + 6

R

t

= 12 Ω

Now the equivalent of the is in parallel to R

3

hence:-

1

𝑅

𝑇

=

1

𝑅

𝑡

+

1

𝑅

3

1

𝑅

𝑇

=

1

12

+

1

6

1

𝑅

𝑇

=

1 + 2

12

𝑅

𝑇

=

12

3

Ans(ii)

𝑅

𝑇

= 4 Ω

Q. 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?

Ans.

V = 220V

P = 10 W

Number of such lamps = ?

SMART EDUCATIONS

35 SMART EDUCATIONS

I

T

= 5 A

𝑃 =

𝑉

2

𝑅

10 =

220

2

𝑅

10 =

48400

𝑅

𝑅 =

48400

10

𝑅 = 4840 Ω

So, the resistance of such bulb is 4840 Ω and let number of such bulbs is which are 𝑥

taking 5A of current in total with 220V, so the total resistance of the circuit will be :-

𝑅

𝑇

=

𝑉

𝐼

𝑇

𝑅

𝑇

=

220

5

𝑅

𝑇

= 44 Ω

So, in parallel connection:-

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

......... 𝑥

1

44

=

1

4840

+

1

4840

......... 𝑥

1

44

=

1

4840

× 𝑥

1

44

=

𝑥

4840

𝑥 =

4840

44

𝑥 = 110

Hence, the number of such bulbs is 110 bulbs. Ans

Q. A hot plate of an electric oven connected to a 220 V line has two resistance coils A

and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel.

What are the currents in the three cases?

Ans.

𝑉 = 220 𝑉

𝑅

1

= 24 Ω

𝑅

2

= 24 Ω

SMART EDUCATIONS

36 SMART EDUCATIONS

(separately, in series, or in parallel)

𝐼 = ?

For separately:-

𝐼 =

𝑉

𝑅

𝐼 =

220

24

Ans(i)

𝐼 = 9 . 17 𝐴

For Series:-

R

T

= R

1

+ R

2

R

T

= 24 + 24

R

T

= 48 Ω

So:-

𝐼 =

𝑉

𝑅

𝐼 =

220

48

Ans(ii)

𝐼 = 4 . 58 𝐴

For parallel:-

1

𝑅

𝑇

=

1

𝑅

1

+

1

𝑅

2

1

𝑅

𝑇

=

1

24

+

1

24

1

𝑅

𝑇

=

1 + 1

24

1

𝑅

𝑇

=

2

24

𝑅

𝑇

= 12 Ω

So:-

𝐼 =

𝑉

𝑅

𝐼 =

220

12

Ans(iii)

𝐼 = 18 . 33 𝐴

SMART EDUCATIONS

37 SMART EDUCATIONS

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

and 2 Ω resistors.

Ans.

Compare P

1

and P

2

for 2

Ω

For P

1

In series:-

V = 6 V

R

1

Ω

R

2

= 2

Ω

For series:-

𝑅

𝑇

= 𝑅

1

+ 𝑅

2

𝑅

𝑇

= 1 + 2

𝑅

𝑇

= 3 Ω

So,

𝐼

𝑇

=

𝑉

𝑅

𝑇

𝐼

𝑇

=

6

3

𝐼

𝑇

= 2 𝐴

In series connection, current is constant hence the power for 2 Ω in series

𝑃 = 𝐼

2

𝑅

𝑃 = 2

2

× 2

𝑃 = 8 𝑊

Hence

𝑃

1

= 8 𝑊

And in parallel connection V is constant hence hence the power for 2 Ω in parallel

𝑃 =

𝑉

2

𝑅

𝑃 =

6

2

𝑃 =

36

2

𝑃 = 18 𝑊

Hence

𝑃

2

= 18 𝑊

Hence P

2

is greater than P

1

Ans

SMART EDUCATIONS

38 SMART EDUCATIONS

Q. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in

parallel to the electric mains supply. What current is drawn from the line if the supply

voltage is 220 V?

Ans

P

1

= 100 W P

2

= 60 W

V

1

= 220 V V

2

= 220 V I

T

= ?

𝑃

1

= 𝑉 𝐼

1

𝑃

2

= 𝑉 𝐼

2

100 = 220 × 𝐼

1

60 = 200 × 𝐼

2

𝐼

1

=

100

220

𝐼

1

=

60

220

𝐼

1

=

5

11

𝐼

1

=

3

11

Hence, the total current in the circuit :-

𝐼

𝑇

= 𝐼

1

+ 𝐼

2

𝐼

𝑇

=

5

11

+

3

11

𝐼

𝑇

=

5 + 3

11

𝐼

𝑇

=

8

11

Ans

𝐼

𝑇

= 0 . 72 𝐴

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

Ans.

Compare E

1

and E

2

P

1

= 250 W P

1

= 1200 W

t = 1 hr t = 10 min

= =

1 × 3600 𝑠 10 × 60 𝑠

𝐸

1

= 𝑃

1

𝑡 𝐸

2

= 𝑃

2

𝑡

𝐸

1

= 250 × 3600 𝐸

2

= 1200 × 600

𝐸

1

= 900000 𝐽 𝐸

2

= 720000 𝐽

Hence the TV uses more energy than the toaster. Ans

Q. 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.

Ans.

SMART EDUCATIONS

39 SMART EDUCATIONS

R = 8 Ω

I = 15 A

t = 2 hrs

the rate at which heat is developed = P = ?

𝑃 = 𝐼

2

𝑅

𝑃 = 15

2

× 8

𝑃 = 225 × 8

Ans

𝑃 = 1800 𝑊

Q . Explain the following.

1. Why is tungsten used almost exclusively for filament of electric lamps?

Ans. For filament of electric lamps a conductor is needed such that it has high

resistance which would make the conductor glow due to high resistance

and has high melting point which would prevent the conductor from

melting in such a high temperature. Tungsten has both the properties:- (i)

high resistance and (ii) high melting point. So tungsten is used almost

exclusively for filament of electric lamps

2. Why are the conductors of electric heating devices, such as bread-toasters and

electric irons, made of an alloy rather than a pure metal?

Ans. The function of all heating devices is to produce heat. According to Joule’s

law of heating to produce more heat, a high resistance conductor is needed and

alloy has higher resistance than pure metal. So, the conductors of electric

heating devices, such as bread-toasters and electric irons, are made of an alloy

rather than a pure metal?

3. Why is the series arrangement not used for domestic circuits?

Ans. In a series arrangement the total resistance increases and it produces more

heat and because of it the devices do not work properly and in addition to this in

a series arrangement more energy is consumed so it is not economical.

4. How does the resistance of a wire vary with its area of cross-section?

Ans. Resistance of a wire is inversely proportional to the area of the

cross-section of a wire.

5. Why are copper and aluminium wires usually employed for electricity

transmission?

Ans. Because they have low resistivity which makes them good conductors of

electricity.