Potential difference between the points $P$ and Q is nearly
Potential difference between the points $P$ and Q is nearly
$17 \mathrm{~V}$
$14 \mathrm{~V}$
$12 \mathrm{~V}$
$8 \mathrm{~V}$
Solution
Let total current be denoted as I. The given circuit is a Wheatstone bridge.
$\begin{aligned}
& \Rightarrow R_1=6+3=9 \Omega . \\
& \Rightarrow R_2=8+4=12 \Omega
\end{aligned}$
According to $\mathrm{KCL}$, the current will get divided into two parts $I_1$ and $I_2$
$\therefore \quad \mathrm{I}_1=\frac{\mathrm{R}_2 \mathrm{I}}{\left(\mathrm{R}_1+\mathrm{R}_2\right)}$
Substituting the values,
$\begin{aligned}
& I_1=\frac{12}{9+12} \times 5 \\
& I_1=2.85 \mathrm{~A}
\end{aligned}$
Potential difference between $P$ and $Q$ is
$\begin{aligned}
& V=I_1 R \\
& V=2.85 \times 6 \\
& V=17 V
\end{aligned}$
