If the heat produced in the 5 , Omega resisto | vedclass

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(B) Let the current in the lower branch (containing $5 \, \Omega$ resistor) be $i_1$ and the current in the upper branch (containing $4 \, \Omega$ and

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the current in the lower branch (containing $5 \, \Omega$ resistor) be $i_1$ and the current in the upper branch (containing $4 \, \Omega$ and $6 \, \Omega$ resistors in series) be $i_2$.Since the branches are in parallel, the potential difference across them is the same: $V = i_1 R_1 = i_2 R_2$.Here, $R_1 = 5 \, \Omega$ and $R_2 = 4 + 6 = 10 \, \Omega$.So, $5 i_1 = 10 i_2 \implies i_1 = 2 i_2$.The heat produced per second (power) is $H = i^2 R$.For the $5 \, \Omega$ resistor: $H_1 = i_1^2 	imes 5 = 10 \, 	ext{cal/sec}$.Substituting $i_1 = 2 i_2$: $(2 i_2)^2 	imes 5 = 10 \implies 20 i_2^2 = 10 \implies i_2^2 = 0.5$.For the $4 \, \Omega$ resistor: $H_2 = i_2^2 	imes 4 = 0.5 	imes 4 = 2 \, 	ext{cal/sec}$.

If the heat produced in the $5 \, \Omega$ resistor is $10 \, \text{cal/sec}$, then how much heat in $\text{cal/sec}$ is produced in the $4 \, \Omega$ resistor?

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