CBSE Class 10 Science 2019 · Paper 31/3/1 · Question 20(a)

Electricity 5 marks Medium

(a) Three resistors of resistances

R_{1}

R_{2}

and

R_{3}

are connected

(i)in series, and
(ii)in parallel. Write expressions for the equivalent resistance of the combination in each case. (b) Two identical resistors of $12 Ω$ each are connected to a battery of $3$ V. Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.

Answer

Series: $R_{s} = R_{1} + R_{2} + R_{3}$ ; parallel: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ ; the power ratio, minimum to maximum resistance, is $4 : 1$ .

(a) (i) In series:

R_{s} = R_{1} + R_{2} + R_{3}

. (ii) In parallel:

\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

. (b) Maximum resistance (series):

R_{s} = 12 + 12 = 24 Ω

. Minimum resistance (parallel):

\frac{1}{R _{p}} = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}

, so

R_{p} = 6 Ω

. With the same

V = 3

P = \frac{V ^{2}}{R}

gives

P_{1} = \frac{9}{6} = 1.5

W for the minimum-resistance combination and

P_{2} = \frac{9}{24} = 0.375

W for the maximum-resistance combination. Ratio

P_{1} : P_{2} = 1.5 : 0.375 = 4 : 1

🔒 Step-by-step solution

CBSE Class 10 Science 2019 · Paper 31/3/1 · Question 20(a)

Answer

From the board papers

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