In the diagram shown,the reading of the voltm | vedclass

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(C) The voltmeter is connected in parallel to the combination of the ammeter and the resistor $R$. Let the resistance of the ammeter be $r_A$. Since t

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Less than $5\,\Omega $

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(C) The voltmeter is connected in parallel to the combination of the ammeter and the resistor $R$. Let the resistance of the ammeter be $r_A$. Since the ammeter and resistor $R$ are in series,the total resistance of this branch is $(R + r_A)$. The voltage across this branch is given by the voltmeter reading,$V = 20\, V$. The current flowing through this branch is given by the ammeter reading,$I = 4\, A$. According to Ohm's law,$V = I(R + r_A)$. Substituting the given values,$20 = 4(R + r_A)$. Dividing by $4$,we get $R + r_A = 5\,\Omega$. Since the ammeter is not ideal,it must have some finite positive resistance,$r_A > 0$. Therefore,$R = 5 - r_A$. Since $r_A > 0$,it follows that $R < 5\,\Omega$.

In the diagram shown,the reading of the voltmeter is $20\, V$ and that of the ammeter is $4\, A$. The value of $R$ should be (Consider the given ammeter and voltmeter are not ideal).

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