The ammeter reading in the circuit below is . | vedclass

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(C) Let the total current be $I = 11 \, A$. The circuit consists of three resistors in parallel: $R_1 = 2 \, \Omega$,$R_2 = 1 \, \Omega$ (with an amme

Vedclass · Accepted Answer

$6$

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(C) Let the total current be $I = 11 \, A$. The circuit consists of three resistors in parallel: $R_1 = 2 \, \Omega$,$R_2 = 1 \, \Omega$ (with an ammeter),and $R_3 = 3 \, \Omega$.The equivalent resistance $R_{eq}$ of the parallel combination is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{2} + \frac{1}{1} + \frac{1}{3} = \frac{3 + 6 + 2}{6} = \frac{11}{6} \, \Omega^{-1}$.Thus,$R_{eq} = \frac{6}{11} \, \Omega$.The potential difference $V$ across the parallel branches is:$V = I 	imes R_{eq} = 11 \, A 	imes \frac{6}{11} \, \Omega = 6 \, V$.The current $I_A$ through the branch containing the ammeter (which has a resistance of $1 \, \Omega$) is:$I_A = \frac{V}{R_2} = \frac{6 \, V}{1 \, \Omega} = 6 \, A$.

The ammeter reading in the circuit below is .............. $A$.

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