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(C) Let the internal resistance of the cell be $r$. Initially,the total resistance of the circuit is $R_{eq} = R_V + r$,where $R_V$ is the resistance

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$A$ will increase,$V$ will decrease.

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(C) Let the internal resistance of the cell be $r$. Initially,the total resistance of the circuit is $R_{eq} = R_V + r$,where $R_V$ is the resistance of the voltmeter. The current is $I = \frac{\varepsilon}{R_V + r}$.When a resistor $R$ is connected in parallel with the voltmeter,the equivalent resistance of the parallel combination becomes $R_p = \frac{R_V R}{R_V + R}$.Since $R_p As the total resistance decreases,the total current $I' = \frac{\varepsilon}{R_p + r}$ increases,so the ammeter reading $A$ increases.The voltage across the parallel combination is $V' = I' R_p = \frac{\varepsilon R_p}{R_p + r} = \frac{\varepsilon}{\frac{R_p + r}{R_p}} = \frac{\varepsilon}{1 + \frac{r}{R_p}}$.Since $R_p$ decreases,the term $\frac{r}{R_p}$ increases,which means the denominator $(1 + \frac{r}{R_p})$ increases,and thus the voltmeter reading $V'$ decreases.

An ammeter and a voltmeter are connected in series with a cell. Their readings are $A$ and $V$ respectively. If a resistor is now connected in parallel with the voltmeter,then:

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