Question 532: Current Electricity

Question

Potential difference between the points P and Q is nearly
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lppdty6f/c0933f30-3d14-4a4c-bd2d-13c1dae0a902/b6364570-91cc-11ee-a087-99308d0b166a/file-6y3zli1lppdty6g.png?format=png)

Accepted Answer

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

Answer

17 V

Answer

14 V

Answer

12 V

Answer

8 V

Current Electricity

Potential difference between the points P and Q is nearly

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