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Potential difference between the points $\mathrm{P}$ and $\mathrm{Q}$ is nearly
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Verified Answer
The correct answer is:
$17 \mathrm{~V}$
As, the resistors across each branch are in series.
$$
\begin{aligned}
& \mathrm{R}_1=6+3=9 \Omega \\
& \mathrm{R}_2=8+4=12 \Omega
\end{aligned}
$$
According to $\mathrm{KCL}$, the current (i) will get divided inte two parts $\mathrm{I}_1$ and $\mathrm{I}_2$
$$
\mathrm{I}_1=\frac{\mathrm{R}_2}{\left(\mathrm{R}_1+\mathrm{R}_2^{\prime}\right) \mathrm{i}}=\frac{12}{9+12} \times 5=2.86 \mathrm{~A}
$$
Potential difference between $\mathrm{P}$ and $\mathrm{Q}$ is $\mathrm{V}=\mathrm{I}_1 \mathrm{R}=2,86 \times 6=17.14 \mathrm{~V}$
$$
\begin{aligned}
& \mathrm{R}_1=6+3=9 \Omega \\
& \mathrm{R}_2=8+4=12 \Omega
\end{aligned}
$$
According to $\mathrm{KCL}$, the current (i) will get divided inte two parts $\mathrm{I}_1$ and $\mathrm{I}_2$
$$
\mathrm{I}_1=\frac{\mathrm{R}_2}{\left(\mathrm{R}_1+\mathrm{R}_2^{\prime}\right) \mathrm{i}}=\frac{12}{9+12} \times 5=2.86 \mathrm{~A}
$$
Potential difference between $\mathrm{P}$ and $\mathrm{Q}$ is $\mathrm{V}=\mathrm{I}_1 \mathrm{R}=2,86 \times 6=17.14 \mathrm{~V}$
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