Question 532: Current Electricity

Question

Potential difference between the points P and Q is nearly

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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