Solution to Question 22
The circuit consists of three parallel branches. Since they are parallel, the total potential difference ($V_{total}$) across each branch is identical.
Each branch contains a resistor $R_i$ and a real voltmeter (resistance $R_V$). The voltmeter reading $V_i$ is the potential drop across $R_V$.
The current in branch $i$ is $I_i = V_i / R_V$.
The total voltage is the sum of the drop across the resistor and the voltmeter:
$$ V_{total} = I_i R_i + V_i = \frac{V_i}{R_V} R_i + V_i = V_i \left( 1 + \frac{R_i}{R_V} \right) $$
Equate the total voltage expressions for branches A and B: $$ V_A \left( 1 + \frac{R_A}{R_V} \right) = V_B \left( 1 + \frac{R_B}{R_V} \right) $$ Given: $V_A = 7.5, R_A = 9; V_B = 5.0, R_B = 14$. $$ 7.5 \left( 1 + \frac{9}{R_V} \right) = 5.0 \left( 1 + \frac{14}{R_V} \right) $$ $$ 1.5 \left( 1 + \frac{9}{R_V} \right) = 1.0 \left( 1 + \frac{14}{R_V} \right) $$ $$ 1.5 + \frac{13.5}{R_V} = 1 + \frac{14}{R_V} $$ $$ 0.5 = \frac{0.5}{R_V} \implies R_V = 1 \, \Omega $$
First, calculate the constant total voltage $V_{total}$ using Branch A values: $$ V_{total} = 7.5 \left( 1 + \frac{9}{1} \right) = 7.5 \times 10 = 75 \text{ V} $$ Now apply the equation to Branch C ($R_C = 24 \, \Omega$): $$ 75 = V_C \left( 1 + \frac{24}{1} \right) $$ $$ 75 = 25 V_C $$ $$ V_C = 3.0 \text{ V} $$