Question 543: Current Electricity

Question

In the following network, the current through galvanometer will
![](https://app-content.cdn.examgoal.net/fly/@width/image/1lt5yreww/074cec53-341a-4708-9e71-8819c40fecbb/bff3cb10-d64f-11ee-a89c-b78526bcb067/file-1lt5yrewx.png?format=png)

Accepted Answer

This is an unbalanced network.
Current through the branch PQR is $\mathrm{I}_1=\frac{\mathrm{V}}{8}$ and current through the branch $\mathrm{PSR}=\mathrm{I}_2=\frac{\mathrm{V}}{4}$
$\begin{aligned}
& 	herefore \mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}=\mathrm{I}_1 	imes 4=\frac{\mathrm{V}}{8} 	imes 4=\frac{\mathrm{V}}{2} \
& 	ext { and } \mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{S}}=\mathrm{I}_2 	imes 1=\frac{\mathrm{V}}{4} 	imes 1=\frac{\mathrm{V}}{4} \
& 	herefore\left(\mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}ight)-\left(\mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{S}}ight)=\frac{\mathrm{V}}{2}-\frac{\mathrm{V}}{4}=\frac{\mathrm{V}}{4} \
& 	herefore \mathrm{V}_{\mathrm{S}}-\mathrm{V}_{\mathrm{Q}}=\frac{\mathrm{V}}{4} \
& 	herefore \mathrm{V}_{\mathrm{S}}>\mathrm{V}_{\mathrm{Q}}
\end{aligned}$
$	herefore$ The current will flow from $\mathbf{S}$ to $\mathbf{Q}$.

Answer

be zero

Answer

flow from $\mathrm{Q}$ to $\mathrm{S}$

Answer

flow in a direction which depends on value of $\mathrm{V}$

Answer

flow from $\mathrm{S}$ to $\mathrm{Q}$

Current Electricity

In the following network, the current through galvanometer will

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