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Two resistors each of resistance 2 ohms are connected in turns (i) series and (ii) in parallel to

a battery of $12 volts$ . Which of the following is the ratio of power consumed in the two cases?

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2 : 3

1 : 4

4 : 2

3 : 4

answer is B.

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Detailed Solution

Two resistors each of resistance 2 ohms are connected in turns (i) series and (ii) in parallel to
a battery of

12 volts

. The ratio of power consumed in two cases is

1 : 4

Given:

R_{1} = R_{2} = 2 ohm

Potential Difference

(V) = 12 V

Considering the first case in which the two resistors are connected in series, the resultant
resistance of the connection is given as
In series
Total Resistance in Series

= R₁ + R₂

= 2 + 2

= 4 ohm

Potential Difference

(V) = 12 V

Current,

(I) = \frac{V}{I}

= \frac{12}{4}

= 3 Ampere

Power

= V \times I

= 12 \times 3

= 36 Watts

Considering the second case in which the two resistors are connected in parallel, the resultant
resistance of the connection is given as,
In parallel

\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

\Rightarrow \frac{1}{2} + \frac{1}{2} = \frac{1}{R_{3}}

\Rightarrow \frac{2}{2} = R_{3}

\Rightarrow R_{3} = 1 ohm

Voltage

(V) = 12 Volts

Current (I) = \frac{V}{R} = 12 Ampere

Power = V \times I

= 12 \times 12

= 144 Watt

Power in series: Power in Parallel

36 : 144

= 1 : 4

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