The current passing through the ideal ammeter | vedclass

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(B) In the given circuit, the two $2 \Omega$ resistors are connected in parallel.Their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{

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(B) In the given circuit, the two $2 \Omega$ resistors are connected in parallel.Their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = 1 \Omega^{-1}$, so $R_p = 1 \Omega$.An ideal ammeter has zero resistance and is connected in parallel with the $4 \Omega$ resistor. This effectively short-circuits the $4 \Omega$ resistor, meaning no current flows through it.The total resistance of the circuit is the sum of the internal resistance of the battery $(1 \Omega)$, the series resistor $(2 \Omega)$, and the parallel combination $(1 \Omega)$.$R_{net} = 1 \Omega + 2 \Omega + 1 \Omega = 4 \Omega$.The total current $I$ flowing from the battery is $I = \frac{V}{R_{net}} = \frac{4 	ext{ V}}{4 \Omega} = 1 	ext{ A}$.Since the ammeter is in parallel with the $4 \Omega$ resistor, all the current $I$ passes through the ideal ammeter.

The current passing through the ideal ammeter in the circuit given below is (in $\text{ A}$)

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