As shown in the circuit diagram,find the valu | vedclass

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(B) For a balanced Wheatstone bridge,the condition is $\frac{R_1}{R_3} = \frac{R_2}{R_4}$. Here,$\frac{2}{4} = \frac{4}{8} = \frac{1}{2}$.Since the br

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$2.5$

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(B) For a balanced Wheatstone bridge,the condition is $\frac{R_1}{R_3} = \frac{R_2}{R_4}$. Here,$\frac{2}{4} = \frac{4}{8} = \frac{1}{2}$.Since the bridge is balanced,no current flows through the galvanometer.The circuit simplifies to two parallel branches:Branch $1$: $R_{s1} = R_1 + R_2 = 2 \Omega + 4 \Omega = 6 \Omega$.Branch $2$: $R_{s2} = R_3 + R_4 = 4 \Omega + 8 \Omega = 12 \Omega$.The equivalent resistance $R_{eq}$ of the parallel combination is:$R_{eq} = \frac{R_{s1} 	imes R_{s2}}{R_{s1} + R_{s2}} = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4 \Omega$.The total current $I$ flowing through the circuit is:$I = \frac{V}{R_{eq}} = \frac{10 	ext{ V}}{4 \Omega} = 2.5 	ext{ A}$.

As shown in the circuit diagram,find the value of $I = \text{ . . . . . . }$. (in $\text{ A}$)

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