The potential difference between the points A | vedclass

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(C) The given circuit is a Wheatstone bridge. The resistances are in the ratio $\frac{8}{10} = \frac{4}{5}$,which means the bridge is balanced. Theref

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$18$

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(C) The given circuit is a Wheatstone bridge. The resistances are in the ratio $\frac{8}{10} = \frac{4}{5}$,which means the bridge is balanced. Therefore,no current flows through the galvanometer,and the resistances in the upper and lower arms are in parallel.The effective resistance $R_{	ext{eff}}$ of the circuit is calculated as:$\frac{1}{R_{	ext{eff}}} = \frac{1}{8+4} + \frac{1}{10+5} = \frac{1}{12} + \frac{1}{15} = \frac{5+4}{60} = \frac{9}{60} = \frac{3}{20} \ \Omega^{-1}$$R_{	ext{eff}} = \frac{20}{3} \ \Omega \approx 6.67 \ \Omega$The total potential difference $V_{AC}$ across the circuit is:$V_{AC} = I 	imes R_{	ext{eff}} = 4 \ 	ext{A} 	imes \frac{20}{3} \ \Omega = \frac{80}{3} \ 	ext{V} \approx 26.67 \ 	ext{V}$Since the bridge is balanced,the potential at $B$ and $D$ is the same. The current flowing through the upper branch $(A-B-C)$ is:$I_{upper} = \frac{V_{AC}}{R_{AB} + R_{BC}} = \frac{80/3}{8+4} = \frac{80/3}{12} = \frac{80}{36} = \frac{20}{9} \ 	ext{A} \approx 2.22 \ 	ext{A}$The potential difference between $A$ and $B$ is:$V_{AB} = I_{upper} 	imes R_{AB} = \frac{20}{9} \ 	ext{A} 	imes 8 \ \Omega = \frac{160}{9} \ 	ext{V} \approx 17.78 \ 	ext{V}$Rounding to the nearest integer,we get $V_{AB} \approx 18 \ 	ext{V}$.

The potential difference between the points $A$ and $B$ is nearly: (in $V$)

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