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Two identical resistors of $2 Ω$ are connected in parallel to a $12 V$ battery. The power consumed by the resistors will be

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12 W

24 W

144 W

132 W

answer is C.

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Detailed Solution

The power consumed by the resistors will be

144 W

.
Given: two identical resistances

R_{1} = R_{2} = 2 Ω

and potential difference

V = 12 V

It is known that “in a parallel combination, the reciprocal of the equivalent resistance is equal to the sum of the reciprocal of the individual resistances.” Mathematically,

\frac{1}{R_{P}} = \frac{1}{R_{!}} + \frac{1}{R_{2}} + . . . + \frac{1}{R_{n}}

Where

R

is the parallel equivalent resistances and

R_{1}, R_{2}, . . ., R_{n}

are the individual resistances.
For two resistances, the equivalent resistance in the parallel combination is given by

\frac{1}{R} = \frac{1}{R_{!}} + \frac{1}{R_{2}}

On substituting the given resistances, we get the parallel equivalent resistance as

\Rightarrow \frac{1}{R} = \frac{1}{2} + \frac{1}{2}

\Rightarrow R = 1 Ω

It is known that the power is given by the formula

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

Where

P

V

and

R

are the power, potential difference and equivalent resistance respectively.
On substituting the given values, we get the power as

\Rightarrow P = \frac{12^{2}}{1}

\Rightarrow P = 144 W

Hence, the power consumed by the resistors will be

144 W

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