If the ammeter in the given figure shows a re | vedclass

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(A) The resistors $3 \, \Omega$ and $6 \, \Omega$ are connected in parallel. Their equivalent resistance $R_p$ is given by:$R_p = \frac{3 	imes 6}{3

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(A) The resistors $3 \, \Omega$ and $6 \, \Omega$ are connected in parallel. Their equivalent resistance $R_p$ is given by:$R_p = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2 \, \Omega$This equivalent resistance $R_p$ is in series with the resistor $R$. The total resistance of the circuit is $R_{eq} = R_p + R = 2 + R$.According to Ohm's law,the total current $I$ in the circuit is given by $I = \frac{V}{R_{eq}}$.Given $I = 2 \, A$ and $V = 6 \, V$,we have:$2 = \frac{6}{2 + R}$$2 + R = \frac{6}{2} = 3$$R = 3 - 2 = 1 \, \Omega$

If the ammeter in the given figure shows a reading of $2 \, A$,find the value of $R$ in $\Omega$.

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