In the circuit as shown in the figure, the he | vedclass

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(D) The circuit consists of two parallel branches connected across the same potential difference $V$.Resistance of the upper branch, $R_1 = 2\,\Omega

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) The circuit consists of two parallel branches connected across the same potential difference $V$.Resistance of the upper branch, $R_1 = 2\,\Omega + 3\,\Omega = 5\,\Omega$.Resistance of the lower branch, $R_2 = 6\,\Omega + 4\,\Omega = 10\,\Omega$.Since the branches are in parallel, the potential difference $V$ across both is the same.Let $i_1$ be the current in the upper branch and $i_2$ be the current in the lower branch.$V = i_1 R_1 = i_2 R_2 \implies i_1(5) = i_2(10) \implies i_1 = 2i_2$.The heat generated per second (power) in a resistor is given by $P = i^2 R$.Given, heat generated in $6\,\Omega$ resistor is $P_2 = i_2^2 	imes 6 = 60\, 	ext{cal/s}$.Thus, $i_2^2 = 10$.We need to find the heat generated in the $3\,\Omega$ resistor, $P_1 = i_1^2 	imes 3$.Substituting $i_1 = 2i_2$, we get $P_1 = (2i_2)^2 	imes 3 = 4i_2^2 	imes 3 = 12i_2^2$.Substituting $i_2^2 = 10$, we get $P_1 = 12 	imes 10 = 120\, 	ext{cal/s}$.

In the circuit as shown in the figure, the heat produced by $6\,\Omega$ resistance due to current flowing in it is $60$ calorie per second. The heat generated across $3\,\Omega$ resistance per second will be ................. $calorie$.

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