Question 14 Solution

Series circuit with Ammeter (resistance $R_A$), Voltmeter (resistance $R_V$), and Resistor $R$. $$ I = \frac{V_{source}}{R_A + R_V + R} $$ Voltmeter reads potential across itself: $V_{read} = I R_V$.

A capacitor $C$ is connected in parallel with the voltmeter.
The effective impedance of the voltmeter section ($Z_V’$) becomes the parallel combination of $R_V$ and $X_C$. $$ |Z_V’| < R_V $$ Since the impedance of this section decreases, the total impedance of the circuit decreases.

Ammeter (A):
Since total impedance decreases, the total current drawn from the source increases.
$\implies$ Reading of A increases.

Voltmeter (V):
Consider the voltage divider. The voltage across the fixed components ($R_A + R$) is $V_{fixed} = I_{total}(R_A + R)$.
Since $I_{total}$ has increased, $V_{fixed}$ increases.
The remaining voltage for the voltmeter section is $V_{source} – V_{fixed}$. Since $V_{fixed}$ increased, the voltmeter reading must decrease.