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Potential difference between the points $P$ and $Q$ in the circuit shown is
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Verified Answer
The correct answer is:
2.88V
The given circuit diagram is
Equivalent resistance between point $P$ and $Q$ is given as
$$
\begin{aligned}
\frac{1}{R_{P Q}} & =\frac{1}{R_A+R_D}+\frac{1}{3}+\frac{1}{R_B+R_C} \\
& =\frac{1}{2+6}+\frac{1}{3}+\frac{1}{4+12}=\frac{1}{8}+\frac{1}{3}+\frac{1}{16} \\
\Rightarrow \quad \frac{1}{R_{P Q}} & =\frac{25}{48} \Rightarrow R_{P Q}=\frac{48}{25} \Omega
\end{aligned}
$$
$\therefore$ Potential difference between the points $P$ and $Q$ is given as
$$
V_{P Q}=I \cdot R_{P Q}=1.5 \times \frac{48}{25}=2.88 \mathrm{~V}
$$
Equivalent resistance between point $P$ and $Q$ is given as
$$
\begin{aligned}
\frac{1}{R_{P Q}} & =\frac{1}{R_A+R_D}+\frac{1}{3}+\frac{1}{R_B+R_C} \\
& =\frac{1}{2+6}+\frac{1}{3}+\frac{1}{4+12}=\frac{1}{8}+\frac{1}{3}+\frac{1}{16} \\
\Rightarrow \quad \frac{1}{R_{P Q}} & =\frac{25}{48} \Rightarrow R_{P Q}=\frac{48}{25} \Omega
\end{aligned}
$$
$\therefore$ Potential difference between the points $P$ and $Q$ is given as
$$
V_{P Q}=I \cdot R_{P Q}=1.5 \times \frac{48}{25}=2.88 \mathrm{~V}
$$
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