EXPLANATIONS

Objective Questions

8. (d) Resistance between terminals A and B

1. (c) Option (c) represents correct set-up for studying the

dependence of the current on the potential difference across

a resistor because ammeter A is connected in series while

voltmeter V is connected across parallel of resistor R.

2Ω

2. (c) At given voltage, current I is inversely proportional to

resistance, i.e. I

∝

¹. At given voltageIis maximum forR₁,

therefore, R₁ is minimum.

2Ω

= (2 Ω, 2 Ω in series) parallel to 2 Ω

2)||2

Ω||

Ω=

9. (a) As, R_eq

I₂

I₃

V(volt)

At given voltage,

I >I

∴

R <R <R

⇒

R >R >R

3. (c) Resistivity depends on the nature of material and the

⇒

temperature but does not depend on the shape of the

3×6

resistor.

3+6=2!

4. (a) True; Specific resistance of a conductor increases on

10. (b) The filament of electric bulb is made up of tungsten

increasing its temperature.

because it has a very high resistance.

5. (c) In first case,

11. (c), As heat produced, H = I2Rt

RAʹ

Resistivity of the conductor, ρ =

…(i)

Here, R =2 Ω

For AD, H₁

× 2 ×1 =18J

In second case,

RAʹ

For DG, H

(−2)

×2 ×1=8J

Resistivity of the conductor, ρ =

…(ii)

For GJ, H

(1)

×2 ×1=2J

Q For same material, resistivity of the conductor will be

same.

∴ Total amount of heat produced in 3 s is,

H=H +H +H

So, on substituting the value of ρ in Eq. (ii), we have

RA RAʹ

⇒ ʹ=2A

⇒

H =18 + 8 + 2 = 28J

12. (b) Electric power determines the rate at which energy is

6. (b) A material that offers higher resistance as compared to

delivered by a current.

conductors to the flow of electron, so has small number of

free electrons is called a poor conductor.

13. (c) Bulbs are rated presuming that they all are connected

with the same voltage supply. In parallel combination,

7. (a) The maximum resistance is obtained when resistors are

voltage across each bulb is same, so greater the watt, greater

connected in series combination.

the brightness.

Thus equivalent resistance

∴ Brightness (B) of 3 bulbs are as follows

Req =R +R +R +R

B >B >B

100

=2 Ω

⇒

B >B >B

14. (d) Given, power, P =1 kW =1000 W

22. (i)

electric energy

−

Voltage,

V = 220 V

¹ =W

time

Current,

I= ?

1000

(ii)

(b) 1 kilowatt-hour is equal to 3.6

×106

joule.

I= P

4.5A

220

(iii)

(a) AsE ∝ t

Thus,the rating offuse-wire is5A whichisgreater than4.5A.

∴ When the time of operating the heater is doubled, the

−3

energy dissipated is doubled.

15. (a) Given, V =

V, I =100

=100

×10

. A

(iv)

∴ Power, P = VI =

∴

E =P × t = 60 × 60

⇒ P = 75mW

⇒

E = 3600 J

16. (c) As we know, resistance of conductor depends on the

(v)

(d) Given, I = 5A, R = 2 Ω, t = 30 min =1800 s

length of conductor as R ∝ l

∴

= 5 × 5 × 2 ×1800

As the length of wire is more, resistance of conductor is

⇒ E = 90000 = 90kJ

more.

∴A is true but R is false.

Subjective Questions

1. Ohm’s law gives the relationship between current I flowing

17. (a) Copper conducts the current without offering much

in a metallic wire and potential difference V, across its

resistance due to high electrical conductivity. Hence,

terminals. According to this law, the electric current flowing

conducting wires are made of copper.

through a conductor is directly proportional to the potential

∴BothA and Rare true and Risthe correct explanationofA.

difference applied across its ends, providing physical

18. (b) Resistors are connected in series, when they join

conditions remains unchanged.

end-to-end consecutively with each other. In parallel

i.e.

V ∝ Ior V = IR

combination, the total resistance is always less than the least

where R is constant of proportionality called resistance of

resistance in combination.

conductor at a given temperature.

∴ Both A and R are true but R is not the correct explanation

of A.

Unit of resistance is ohm. It is said to be 1 ohm, if potential

difference of 1 volt across the ends of the conductor makes a

19. (b) Fuse is a safety device that is used in household circuits.

current of 1 A to flow through it.

It is connected in series with the main supply.

−1

Fuse consists of an alloy of lead and tin which has

i.e.

1 ohm (Ω) =

appropriate melting point.

2. Resistance is the property of a conductor by virtue of which

∴ Both A and R are true but R is not the correct explanation

it opposes/resists the flow of charges/flow of current through

of A.

it. Its SI unit is ohm and is represented by the Greek letter

Ω (ohm). Resistance of a conductor is given by R

20. (a) The resistance, R

3. Scale : At x-axis, 1 div (1 cm) = 0.1 A

R∝

At y-axis, 1 div (1 cm) = 0.5 V

i.e. Higher the wattage of a bulb, lesser is the resistance and

The graph between current and potential difference is

so it will glow brighter.

shown below.

∴Both A and R are true and R is the correct explanation of A.

5.0

21. (i)

(a) Materials that follow Ohm’s law are called ohmic

4.0

conductor. V - I graph is a straight line passing from

3.0

origin for ohmic conductors.

2.5

(ii)

(Volts)

2.0

voltage axis at room temperature.

1.5

1.0

(iii)

(b) Slope of V -I graph =

=1.

V R

0.5

(iv)

(d) On increasing the voltage, the resistance remains

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

same but current increases.

I (Ampere)

(v)

(a) Given, V = 9V, I = 0.1 A

Resistance (R) of the resistor is determined by the slope of

∴

R = V

V-I graph.

⎡0.25

⎤

∴ Resistance = R=Slope of graph =

−3

5×

×10

⎢

⎥

⎣

⎦

−

1.5−1.0

⇒

−8

∴

R= y

0.8×10

x −x

0.3−0.2

0.5

[Q A = πr² and r

]

0.1

−8

5×

×1.56×10

4. Alloys have a higher resistivity than their constituent metals.

−8

0.8×10

They do not oxidise or burn at higher temperatures as they

ρl

have high melting point. Thus, they are used to make coils of

(ii)

∴Resistance, R =

electrical toasters and electric irons rather than pure metals.

ρl

5. (i) Following factors on which resistance of a wire depends

2 =

⎛D⎞

π D²

(a) length of wire : R∝l

⎜

⎟

⎝

2⎠

(b) area of cross-section of wire : R∝¹

[Q A = πr² and r

]

ρl

R'= ρl=

[Q D has become 2D]

∴

R=ρ l

⎛ 2D⎞

⎜

⎟

⎝

⎠

(ii) Metals are good conductors as their resistivity is very low

whereas glass is a bad conductor having high resistivity.

= ρ

πD²

(iii) Alloys are used as heating elements as their resistivity

and melting points both are very high.

ρl

ρl×4

Now,

6. The given circuit is shown below

πD²

πD

ρl

πD

ρl

∴

R'=R

12 V

R₁=4 Ω

R₂

Thus, resistance will decrease by 4 times.

Let R be the resistance of the entire circuit.

5Ω

8Ω

12 Ω

Given, overall current needed, I = 9 A

R₁

R₂

Voltage,

V =12 V

Using Ohm's law, V = IR

Equivalent resistance of the circuit,

⇒

R = V

R=R +R +R

=5 + 8 + 12 = 25Ω

Now, the resistors R₁ and R₂ are in parallel combination.

[QR , R

2 andR3 areconnectedinseries]

In series combination, current flowing through all the

∴

⇒

R R

resistances is same and equal to the total current flowing

through the circuit.

⇒

[QR =

Ω]

∴ Current in the resistors,

I= V

0.24 A

⇒

−

∴ Ammeter reading = 0.24 A

⇒ R₂

Potential across 12 Ω resistance,

So, the student should choose 2 Ω resistor.

V =IR = 0.24 ×12 = 2.88V

7. Given, diameter = 0.25 mm

∴ Voltmeter reading is 2.88 V.

−8

Resistivity, ρ =

0.8×10

Ω-m

9. Parallel combination of resistances is highly useful in

(i) Resistance, R = 5 Ω

circuits used in daily life, as the circuits used have

ρl

We know that, R =

components of different resistances requiring different

amounts of current.

This type of combination in a circuit divides the current

Now, the given circuit can be redrawn as shown below

among the components (electrical gadgets), so that they can

R₅=3 Ω

have necessary amount of current to operate properly. This

is the reason of connecting electrical appliances in parallel

combination in household circuit.

R''=2 Ω

10. (i) Resistance is maximum when resistors are connected in

series.

4Ω

8Ω

12 Ω

24 Ω

R₁=3 Ω

R_max = 4 + 8 +12 + 24 = 48 Ω

Now, it is clear from the above circuit that all the resistances

R5,Rʹʹ and R1 are in series combination.

(ii) Resistance is minimum when resistors are connected in

parallel.

As, current through R1,Rʹʹ and R5

is same.

4Ω

∴ Equivalent resistance of the circuit is

R = R + Rʹʹ+R

3 Ω+2 Ω+3 Ω=8 Ω

1 =

8Ω

12. The following figure shows the connection of resistors in

parallel

12 Ω

R₁

24 Ω

I₁

R₂

⎡1

⎤

Ω=

I₂

R₃

Rmin. =

⎢

⎥

⎣4

24⎦

I₃

11. From the combination, it can be observed that R₂ and R₃ are

in series order.

R₅

=3Ω

R₃=3Ω

An applied potential difference V produces current I₁ in

R ,I

in R₂ and I₃ in R₃.

Total current, I = I

…(i)

R₁=3Ω

R₂=3Ω

By Ohm’s law, I

and I

As current through R₂ and R₃ is same. So, their equivalent

If R is the equivalent resistance, then I

resistance is Rʹ=R

3Ω+3Ω=6Ω

Now, the given circuit can be redrawn as shown below

Thus,

[from Eq. (i)]

R₅=3Ω

⎛

⎞

⇒ V

V⎜

⎟

⎝

R R R

⎠

⇒

R R

=6Ω

13. The brightness of glow of bulb P will increase and brightness

of glow of bulb Qwill decrease.

R₁=3Ω

This is because, on closing S, bulbs Q and R will be in parallel

Now, it can be seen that R₄ and Rʹ are in parallel

and the combination will be in series with bulb P. Hence, the

combination. As, currents through R₄ and Rʹ are

total resistance of the circuit will decrease and the current

different. So, their equivalent resistance can be calculated

flowing in the circuit will increase. Therefore, the glow of

as below

bulb P will increase.

Also, bulbs Qand R will be in parallel in this case. So, the

Rʹʹ Rʹ R

current gets divided and lesser current flows through Qand

hence the glow of bulb Qdecreases.

∴ Rʹʹ = 2 Ω

14. (i) According to Joule’s law of heating, amount of heat

P = V2/R is used when potential difference(V)across every

produced in a resistor is

component of the circuit is constant. This expression is used

(a) directly proportional to square of current flowing

in case of parallel combination in the circuit. In series

through the resistor.

combination, R is greater than the value of R in parallel

combination.

∴

H∝I²

(b) directly proportional to resistance of the resistor.

20. Resistance, R = 2 Ω

∴

H∝R

Maximum power, P_max =18 W

Maximum current, I_max

flows through the resistor.

P=I2R

∴

H∝t

⇒

3A=I_max

Hence,

H=I2Rt

(ii)

Maximum current that can flow through 2 Ω resistor is 3 A.

L₁

100 W, 220 V

This current divides along B and C because they are in

parallel combination. Voltage across B and C remains same

L₂

60 W, 220 V

and hence I

∝

¹. Since, B and C have same resistance 2 Ω

each, and has same current. i.e. I =

=1.5 A flowing through

220 V

B and C.

21. Given, power, P =1500 W,

Here, potential, V = 220 V

Power, P₁ = 100 W, P₂ = 60 W

Voltage, V = 250 V

As, current drawn is given by

(i) Electric current drawn,

I= Power(

[From P = VI]

1500

Voltage (V)

I= P

250

100

So,

I₁

A and I =

(ii) Energy consumed, E = Power × Time

220

=1500 × 50

[Q t = 50 h]

15. Given, R₁

Ω, R₂

Ω, t = 5s , V = 6 V

= 75000 Wh = 75 kWh

∴Net resistance, R = R

2 Ω+ 4 Ω=6 Ω

[Q 1 kW = 1000 W]

∴ Current, I

=1A

= 75 unit

[Q 1 unit = 1 kWh]

In series, same 1 A current passes through both resistors.

(iii)Q Cost of energy consumed = 75 × 6 = ` 450

∴ Heat dissipated, H = I2R

×t

=(1)

5= 20 J

22. The given circuit is shown below

16. The heating element of heater is made up of an alloy that has

very high resistance. So when the current flows through it, it

becomes very hot and glows red. But the resistance of cord

60 V

is less because it is made up of copper or aluminium, so it

does not glow.

4A B

5A C

3A D

17. Electric fuse is used to protect the electrical circuit from

overloading and short circuit. When the current flowing

through a circuit exceeds the safe limit, the temperature of

fuse wire increases and due to heating effect, it gets melt

and breaks the circuit.

(i) In the circuit, all the lamps have same voltage, i.e. 60 V

18. The SI unit of electric power is watt (W). It is the power

but each lamp is having different current. So, the lamps

consumed by a device that carries 1 A of current when

are arranged in parallel combination.

operated at a potential difference of 1 V. Thus,

(ii) The two advantages of lamps in parallel combination are

1W =1 volt ×1 ampere =1 VA

(a) if one lamp gets faulty, it will not affect the working

The commercial unit of electrical energy is kilowatt hour

of other lamps.

(kWh), commonly known as ‘unit’.

(b) in parallel combination of lamps, each lamp will use

1 kWh =

the full potential of the battery.

19. P = I2R is used when current flowing in every component of

(iii) The lamp with the highest power will glow the brightest.

the circuit is constant. This is the case of series combination

As, power = Voltage × Current

of the devices in the circuit.

In this case, all the lamps have same voltage i.e., 60 V.

For lamp A , current = 3 A

Potential drop at R₁ ,

∴

Power = 60 × 3 = 180 W

V =IR

=2 ×2 =4V

For lamp B, current = 4 A

Potential drop at R₅ ,

∴

Power = 60 × 4 = 240 W

V =IR

=2 ×2 =4V

For lamp C, current = 5 A

Now, potential drop at Rʹ , Vʹ can be calculated as

∴

Power = 60 × 5 = 300 W

V = V + V + Vʹ

For lamp D, current = 3 A

⇒

9 = 4 + 4 + Vʹ

∴

Power = 60 × 3 = 180 W

⇒

Vʹ = 1 V

As, the lamp C is having the maximum power, so it will

I₂

R₂

glow the brightest.

(iv) Let R be the total resistance of the circuit.

I₃

Total current in the circuit, I = 3 + 4 + 5 + 3 = 15 A

R₃

I=I₁

R₁

Voltage, V = 60 V

Using Ohm's law, V = IR

I₄

R₄

R₅

⇒

R = V

4Ω

23. In the given circuit, R ,R

and R₄ are in parallel

As R ,R

and R₄ are in parallel combination, so potential

combination. As, currents through R ,R

and R₄ are

drop at all resistances will be same as 1 V.

different. So, their equivalent resistance R is

V =V =V =V

ʹ=

1Ω

Current through R₂ ,

R₂

Vʹ

2Ω

=1A

2Ω

3Ω

2Ω

Vʹ

Similarly,

05A

R₄

R₅

–

Vʹ

and

033

24. (i) A states that the electric current flowing through a

Rʹ

conductor directly proportional to the potential

difference applied across its ends provided the physical

conditions such as temperature remains unchanged.

(ii) An ammeter is connected in series in a circuit.

⇒

Rʹ =

(iii) Given, P =100 W and time t =1 min = 60 s.

As, energy E = Pt =100 × 60 = 6000 J

Now, the given circuit can be redrawn as shown below

(iv) Given, a wire with resistance 5 Ω.

2Ω

6/11 Ω

2Ω

Now, wire is converted into a ring as shown below

R₁

Rʹ

R₅

Now, R1,Rʹ and R5 are in series combination. As, current

through R ,Rʹ and

is same.

The equivalent circuit,

So, equivalent resistance of the whole circuit is

R = R + Rʹ + R

22+

=2+

2 =

Now, total current flowing through the circuit,

2.5×

2.5

Hence, R =R

1.25W

2.5+

2.5

I= V

≈

25. Given, resistance of lamp (R )

= 20 Ω

Resistance of conductor (R )

Current through R₁ and R₅ will be same as these are in

Potential difference of battery (V) = 6 V

series combination and will be equal to the total current

(i) Total resistance,

flowing through the circuit.

R=R +R

2 =

20 + 4 = 24 Ω

∴

I=I

2 A

(ii) Current through the circuit

(ii) Resistance of bulb,

I= V

R= V

(Ohm’s law)

(iii) (a) Potential difference across electric lamp = IR₁

⇒ R=

=10Ω

−

× 20 = 5 V

500

×10

(b) Potential difference across conductor

(iii) Energy consumed in 2¹ hour

= IR₂ = 0.25 × 4 = 1 V

(iv) Power of lamp = I2R₁

2.5×

2.5

P.t.=

[2¹ hour = 2.5 hour]

1000

=(0.25

)

6.25

= 0.00625 kWh

1000

26. (i) Power is given as, P = VI where V is the voltage and I is

28. (i)

The electrical circuit in which some resistances are

the current. Power consumed is minimum when the

connected in series combination and some in parallel

current passing is minimum. So, the resistors should be

combination, this type of combination is called complex

connected in series.

circuit.

(ii) Power of each bulb=100 W

(ii)

Resistance in series is always greater than resistance in

Total power of 3 bulbs

parallel. The slope of V-I graph gives resistance.

PB =

3×100 = 300 W

Hence, line B denotes resistance in series combination

Energy consumed by 3 bulbs in a day,

and line A denotes resistance in parallel combination.

E =P ×t

B B

(iii)

Equivalent circuit diagram is

= 300W × 5 h =1.5 kWh

[Qt_B = 5h]

Similarly, power of 2 fans, P_F = 2 × 50 W =100 W

Energy consumed by 2 fans, E =P ×t

F F

= 100 W ×10 h

[Qt_F =10h]

=1kWh

Now, energy consumed by heater in a day

EH H × H=

R_S

=1kWh

×¹h

∴P_H =1kWh, t_H =

R = R +R =2R

^S1

=2R +2R =4R

= 0.5kWh

^S2

Both of them are in parallel.

Total energy consumed in a day

R ×R

E =E +E +E

∴

eq =

R +R

=(1.5

05)

kWh = 3 kWh

Energy consumed in a month of 31 days

(iv) When the equivalent resistance is less than the resistance

of least resistor. This is known as minimum effective

= E × 31= 93kWh

resistance.

∴The cost of energy consumed

(v) In series combination, the value of current is same

= (` 3.60/kWh) × (93 kWh)

through each resistor.

=`334.80

So, it will not be affected at all.

27.(a) Power is defined as the rate at which electric energy is

29. (i) According to Joule’s law of heating, the heat produced in

dissipated or consumed in an electric circuit.

a resistor is directly proportional to the square of current

²/R

flowing through that resistor.

The SI unit of electric power is watt (W).

(ii) Practical application of heating effect of current is

It is the power consumed by a device that carries 1 A of

electric heater, electric iron etc.

current when operated at a potential difference of 1 V.

(iii) Electrical energy is transformed into heat energy.

1 W=1volt×1 ampere=1VA

(iv) Given, resistance R = 25 Ω; current I = 7 A;

(b) Given, voltage rating, V = 5 V and current rating I= 500 mA

Time t = 0.5 min = 0.5 × 60 = 30 s. Heat H = ?

(i) As, we know power of bulb

We know that, Heat H = I2Rt

−3

P = VI =

500

×10

[Q A

H = (7)

× 25 × 30 = 36750J

−

1mA =10

³A]

= 368×10⁴

= 2.5 W

−

So, the heat developed is 3.68

×104

2.5×

³ kW

(v) Given, Heat H = 200 J, Resistance, R = 5 Ω

(iii)

Time t = 10 s, Potential difference V = ?

⇒

W = P × T= VIT

(iv)

(b) Resistance of each resistor =

. Ω

We know that,

Total resistance in series R

S =

n×R

= 200 ×

Heat, H = I2Rt

⇒

R_S = 40 Ω

(v)

(d) Resistor of 6 Ω and 2 Ω are in parallel.

⇒ Current, I

R ×R

∴ R

P =

200

7 =

5×10

So, the potential difference across the resistor is

V = IR

[by Ohm’s law]

R_P = Ω

= 2 × 5 =10V

Equivalent circuit is shown below,

30. (i)

(a) An applied potential V produces current I in the

3/2 Ω

6Ω

resistors and R ,R

and R₃ causing a potential drop

V ,V

and V₃ respectively, through each resistor.

Both of them are in series

Total potential, V = V

+ V + V

⎛

⎞

(ii)

(a) In series combination of resistors, the current is same

∴

R_eq =

⎜

6⎟ Ω

⎝2

⎠

at every point of the circuit, i.e. the current through each

resistor is same.

⇒

R_eq = Ω