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In the circuit shown in the figure the heat produced in the 5ohm resistor due to the current flowing through it is 10cal/sec. The heat produced in 4ohm resistor, in cal/sec, is ___________

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detailed solution

Correct option is B

$\begin{array}{l} P = \frac{V^{2}}{R} \Rightarrow \frac{V^{2}}{5} = 10 \to (1) \\ V_{1} + V_{2} = V \to (2) \\ \frac{V_{1}}{4} = \frac{V_{2}}{6} \\ \Rightarrow 3 V_{1} = 2 V_{2} \to (3) \\ From (2) & (1) \\ \Rightarrow 5 V_{1} = 2 V \Rightarrow V_{1} = \frac{2 V}{5} \end{array}$
$so the heat produced in the 4 Ω resistor, P = \frac{V_{1}^{2}}{4} = \frac{4 V^{2}}{25} \times \frac{1}{4} (from (1) V^{2} = 10 \times 5 = 50 {V o l t}^{2}) \Rightarrow P = \frac{V^{2}}{25} = \frac{1}{25} \times 50 = 2 c a l / s e c$

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detailed solution

Correct answer is 2

$\begin{array}{l} P = \frac{V^{2}}{R} \Rightarrow \frac{V^{2}}{5} = 10 \to (1) \\ V_{1} + V_{2} = V \to (2) \\ \frac{V_{1}}{4} = \frac{V_{2}}{6} \\ \Rightarrow 3 V_{1} = 2 V_{2} \to (3) \\ From (2) & (1) \\ \Rightarrow 5 V_{1} = 2 V \Rightarrow V_{1} = \frac{2 V}{5} \end{array}$
$so the heat produced in the 4 Ω resistor, P = \frac{V_{1}^{2}}{4} = \frac{4 V^{2}}{25} \times \frac{1}{4} (from (1) V^{2} = 10 \times 5 = 50 {V o l t}^{2}) \Rightarrow P = \frac{V^{2}}{25} = \frac{1}{25} \times 50 = 2 c a l / s e c$

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