In the circuit shown in the figure, the heat | vedclass

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(B) The circuit consists of two parallel branches. The upper branch has a total resistance of $R_2 = 4\, \Omega + 6\, \Omega = 10\, \Omega$. The lower

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(B) The circuit consists of two parallel branches. The upper branch has a total resistance of $R_2 = 4\, \Omega + 6\, \Omega = 10\, \Omega$. The lower branch has a resistance of $R_1 = 5\, \Omega$.Since the branches are in parallel, the potential difference $V$ across both branches is the same.The current in the branches is inversely proportional to their resistance: $\frac{i_1}{i_2} = \frac{R_2}{R_1} = \frac{10}{5} = \frac{2}{1}$.The heat produced per second (power) is given by $P = i^2 R$.For the $5\, \Omega$ resistor, $P_5 = i_1^2 	imes 5 = 10\, 	ext{cal/s}$.For the $4\, \Omega$ resistor, the current is $i_2$. The power dissipated in the upper branch is $P_{	ext{upper}} = i_2^2 	imes R_2 = i_2^2 	imes 10$.Using the ratio $\frac{i_1}{i_2} = 2$, we have $i_1 = 2i_2$. Substituting this into the power equation for the $5\, \Omega$ resistor: $(2i_2)^2 	imes 5 = 10 \Rightarrow 20i_2^2 = 10 \Rightarrow i_2^2 = 0.5$.The power dissipated in the $4\, \Omega$ resistor is $P_4 = i_2^2 	imes 4 = 0.5 	imes 4 = 2\, 	ext{cal/s}$.

In the circuit shown in the figure, the heat produced in the $5\, \Omega$ resistance is $10\, \text{cal/sec}$. The heat produced in the $4\, \Omega$ resistance is ................ $\text{cal/sec}$.

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