Q.The heat developed across 6 Ω resistor per second is 50 J. Calculate the heat developed per second across 2 Ω resistor in the given electric circuit. [figure: circuit diagram — current I enters a node that splits into two parallel branches between two nodes: one branch has a 2 Ω resistor in series with a 3 Ω resistor, the other branch has a single 6 Ω resistor; the branches rejoin and current I exits, as printed]

Since the $6\,\Omega$ resistor and the $(2+3)\,\Omega$ series branch are connected in parallel across the same two nodes, they have the same voltage across them; working from the given $6\,\Omega$ heat rate gives the branch voltage, and then the heat rate in the $2\,\Omega$ resistor.

From the circuit, current $I$ enters a node and splits into two parallel paths that reconnect at a second node before the current $I$ leaves: one path is a single $6\,\Omega$ resistor, and the other path is a $2\,\Omega$ resistor in series with a $3\,\Omega$ resistor (total $5\,\Omega$). Because both paths connect the same pair of nodes, the potential difference $V$ across the $6\,\Omega$ resistor equals the potential difference across the $5\,\Omega$ series branch.

Step 1 — find $V$ from the $6\,\Omega$ branch. The heat (power) dissipated per second in a resistor is $P = V^2/R$. Given that the $6\,\Omega$ resistor dissipates $50\,\text{J}$ per second,

$P_6 = \dfrac{V^2}{6} = 50 \quad\Rightarrow\quad V^2 = 300\ \text{V}^2$

Step 2 — find the total power in the $5\,\Omega$ branch. The same voltage $V$ appears across the series combination of $2\,\Omega$ and $3\,\Omega$ (total $5\,\Omega$), so the total power dissipated in that branch is

$P_{5} = \dfrac{V^2}{5} = \dfrac{300}{5} = 60\ \text{W}$

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