The potential difference between the points P | vedclass

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Question

(A) Let the total current entering at $P$ be $I = 5 \,A$. The circuit is a bridge network. Let the potential at $P$ be $V_P$ and at the output node be

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) Let the total current entering at $P$ be $I = 5 \,A$. The circuit is a bridge network. Let the potential at $P$ be $V_P$ and at the output node be $V_R = 0 \,V$.
We need to find the potential at $Q$ $(V_Q)$.
The upper branch has resistors $6 \,\Omega$ and $3 \,\Omega$ in series. The lower branch has resistors $8 \,\Omega$ and $4 \,\Omega$ in series.
Let $I_1$ be the current in the upper branch and $I_2$ be the current in the lower branch.
$I_1 = I 	imes \frac{R_{lower}}{R_{upper} + R_{lower}} = 5 	imes \frac{8+4}{(6+3) + (8+4)} = 5 	imes \frac{12}{9+12} = 5 	imes \frac{12}{21} = 5 	imes \frac{4}{7} \approx 2.857 \,A$.
The potential at $Q$ relative to $P$ is the voltage drop across the $6 \,\Omega$ resistor.
$V_P - V_Q = I_1 	imes 6 = 2.857 	imes 6 = 17.14 \,V$.
Thus,the potential difference between $P$ and $Q$ is approximately $17 \,V$.

The potential difference between the points $P$ and $Q$ is nearly: (in $\,V$)

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