In the circuit shown,the readings of the amme | vedclass

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(C) The measured resistance is $R_{measured} = \frac{V}{I} = \frac{20\,V}{4\,A} = 5\,\Omega$.In the given circuit,the voltmeter is connected in parall

Vedclass · Accepted Answer

greater than $5\,\Omega$

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(C) The measured resistance is $R_{measured} = \frac{V}{I} = \frac{20\,V}{4\,A} = 5\,\Omega$.In the given circuit,the voltmeter is connected in parallel with the resistor $R$.Since the voltmeter is non-ideal,it has a finite resistance $R_V$. Therefore,the total current $I$ measured by the ammeter splits into two parts: current $I_R$ through the resistor $R$ and current $I_V$ through the voltmeter.Thus,$I = I_R + I_V$,which implies $I_R = I - I_V$.The actual resistance $R$ is given by $R = \frac{V}{I_R} = \frac{V}{I - I_V}$.Since $I_V > 0$,it follows that $I - I_V Therefore,$R = \frac{V}{I - I_V} > \frac{V}{I} = 5\,\Omega$.Hence,the actual resistance $R$ is greater than $5\,\Omega$.

In the circuit shown,the readings of the ammeter and voltmeter are $4\,A$ and $20\,V$ respectively. The meters are non-ideal. Then $R$ is:

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