CBSE Class 10 Science 2020 · Paper 31(B) · Question 30(a)

Electricity 5 marks Medium

Two resistors A and B of resistances

10 Ω

and

40 Ω

respectively are first joined in series and then in parallel with two terminals of a battery. In each case the voltage applied is 5 V.

(a)Find the ratio of currents through the resistors A and B (i.e. $I_{A} / I_{B}$ ) in each case.
(b)Find the ratio of voltages across the resistors A and B (i.e. $V_{A} / V_{B}$ ) in each case.
(c)Find the heat produced each second in the series combination of resistors.

Answer

Series: $I_{A} / I_{B} = 1$ , $V_{A} / V_{B} = \frac{1}{4}$ ; parallel: $I_{A} / I_{B} = 4$ , $V_{A} / V_{B} = 1$ ; the series combination produces 0.5 J of heat each second.

(a) Series: the same current,

I = \frac{5}{50} = 0.1

A, flows through both, so

I_{A} / I_{B} = 1

(

1 : 1

). Parallel:

I_{A} = \frac{5}{10} = 0.5

A and

I_{B} = \frac{5}{40} = 0.125

A, so

I_{A} / I_{B} = 4

(

4 : 1

). (b) Series:

V_{A} = 0.1 \times 10 = 1

V and

V_{B} = 0.1 \times 40 = 4

V, so

V_{A} / V_{B} = \frac{1}{4}

(

1 : 4

). Parallel: both are across the battery,

V_{A} = V_{B} = 5

V, so

V_{A} / V_{B} = 1

(

1 : 1

). (c)

H = I^{2} R t = (0.1)^{2} \times 50 \times 1 = 0.5

J each second.

🔒 Step-by-step solution

From the board paper

CBSE Class 10 Science 2020 · Paper 31(B), Question 30(a) — all 30 questions with answers →

CBSE Class 10 Science 2020 · Paper 31(B) · Question 30(a)

Answer

From the board paper

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