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

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(C) The ammeter measures the total current $I$ flowing through the circuit,and the voltmeter measures the potential difference $V$ across the resistor

Vedclass · Accepted Answer

greater than $5\,\, \Omega$

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(C) The ammeter measures the total current $I$ flowing through the circuit,and the voltmeter measures the potential difference $V$ across the resistor $R$. Since the voltmeter is connected in parallel with the resistor $R$,it is non-ideal and 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. So,$I = I_R + I_V$. Since $I_V > 0$,the current through the resistor $I_R = I - I_V$ is less than the total current $I$ measured by the ammeter. According to Ohm's law,the actual resistance $R$ is given by $R = \frac{V}{I_R}$. Since $I_R  \frac{V}{I}$. Given $V = 20\,\, V$ and $I = 4\,\, A$,the measured value is $\frac{V}{I} = \frac{20}{4} = 5\,\, \Omega$. Thus,the actual resistance $R$ must be 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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