The potential difference between the points P | vedclass

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(C) The circuit consists of two parallel branches connected between points $P$ and $R$. The total current entering at $P$ is $I = 5 \, A$.Branch $1$ (

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) The circuit consists of two parallel branches connected between points $P$ and $R$. The total current entering at $P$ is $I = 5 \, A$.Branch $1$ (upper) has resistors $6 \, \Omega$ and $3 \, \Omega$ in series. Total resistance $R_1 = 6 + 3 = 9 \, \Omega$.Branch $2$ (lower) has resistors $8 \, \Omega$ and $4 \, \Omega$ in series. Total resistance $R_2 = 8 + 4 = 12 \, \Omega$.Using the current divider rule, the current $I_1$ flowing through the upper branch is:$I_1 = I 	imes \frac{R_2}{R_1 + R_2} = 5 	imes \frac{12}{9 + 12} = 5 	imes \frac{12}{21} = 5 	imes \frac{4}{7} \approx 2.857 \, A$.The potential at $P$ relative to $Q$ is the voltage drop across the $6 \, \Omega$ resistor:$V_{PQ} = I_1 	imes 6 = 2.857 	imes 6 \approx 17.14 \, V$.Thus, the potential difference is nearly $17 \, V$.

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

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