Question 530: Current Electricity

Question

Potential difference between the points A and B is nearly

Accepted Answer

The given circuit is a Wheatstone's bridge, As the resistances are in the same ratio, the bridge is balanced. Therefore, no current will flow through the galvanometer and the resistances will be in parallel combination.
$$\begin{aligned}
& 	herefore \quad 	ext { Voltage between } A 	ext { and } C 	ext { is } V_{A C}=I . R_{	ext {eff }} \
& \frac{1}{R_{e f f}}=\frac{1}{(8+4)}+\frac{1}{(10+5)} \Rightarrow R_{e f f}=\frac{15 	imes 12}{27}=6.67 \Omega \
& 	herefore \quad V_{A C}=4 	imes 6.67=26.68 \mathrm{~V} \quad \ldots[	ext { From(i) }]
\end{aligned}$$
$$\begin{aligned}
& 	herefore \quad \mathrm{I}_{A C}=\frac{V_{A C}}{R_{A B}+R_{B C}}=\frac{26.68}{8+4}=2.22 \mathrm{~A} \
& \quad V_{A B}=I_{A C} 	imes R_{A B}=2.22 	imes 8=17.76 \mathrm{~V}=18 \mathrm{~V}
\end{aligned}$$

Answer

10 V

Answer

14 V

Answer

18 V

Answer

20 V

Current Electricity

Potential difference between the points A and B is nearly

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