Voltmeter Reading Calculation

An analog voltmeter operates based on Ohm's Law. The voltage reading ($V_{reading}$) is proportional to the current ($I$) flowing through the meter movement, i.e., $V_{reading} = k \cdot I$, where $k$ is a constant. The current $I$ is determined by the applied voltage ($V_{applied}$) and the total resistance ($R_{total}$) of the circuit, which includes the internal resistance of the meter ($R_{int}$) and the external multiplier resistance ($R_m$). Thus, $I = \frac{V_{applied}}{R_{int} + R_m}$.

Combining these, we get: $V_{reading} = k \cdot \frac{V_{applied}}{R_{int} + R_m}$.

This can be rearranged as: $V_{applied} = \frac{V_{reading}}{k} (R_{int} + R_m)$.

For problems like this, it's often assumed that the applied voltage remains constant across the different settings, and the multiplier resistance is changed, causing the meter reading to vary. Let this constant applied voltage be $V_0$. Then:

$V_0 = \frac{V_{reading}}{k} (R_{int} + R_m)$

This implies that the product $V_{reading} \cdot (R_{int} + R_m)$ must be constant for a fixed $V_0$ and $k$. Let this constant be $C$.

$C = V_{reading} \cdot (R_{int} + R_m)$

Determining Internal Resistance ($R_{int}$)

We are given two scenarios:

• Scenario 1: $R_{m1} = 20 \text{ k}\Omega = 20000 \Omega$, $V_1 = 440 \text{ V}$
• Scenario 2: $R_{m2} = 80 \text{ k}\Omega = 80000 \Omega$, $V_2 = 352 \text{ V}$

Using the constant product relationship:

Equation 1: $440 \text{ V} \times (R_{int} + 20000 \Omega) = C$

Equation 2: $352 \text{ V} \times (R_{int} + 80000 \Omega) = C$

Equating the two expressions for $C$:

$440 (R_{int} + 20000) = 352 (R_{int} + 80000)$

$440 R_{int} + 8800000 = 352 R_{int} + 28160000$

Solving for $R_{int}$:

$(440 - 352) R_{int} = 28160000 - 8800000$

$88 R_{int} = 19360000$

$R_{int} = \frac{19360000}{88} \Omega = 220000 \Omega = 220 \text{ k}\Omega$

Calculating the Reading for the New Setting

Now we need to find the reading ($V_3$) when the multiplier setting is $R_{m3} = 40 \text{ k}\Omega = 40000 \Omega$. We first calculate the constant $C$ using one of the known scenarios. Using Scenario 1:

$C = 440 \text{ V} \times (220000 \Omega + 20000 \Omega)$

$C = 440 \text{ V} \times 240000 \Omega = 105600000 \text{ V}\Omega$

Now, use the constant $C$ and the new multiplier setting ($R_{m3}$) to find the reading $V_3$:

$C = V_3 \times (R_{int} + R_{m3})$

$105600000 \text{ V}\Omega = V_3 \times (220000 \Omega + 40000 \Omega)$

$105600000 = V_3 \times 260000$

$V_3 = \frac{105600000}{260000}$

$V_3 = \frac{10560}{26} \text{ V}$

$V_3 = \frac{5280}{13} \text{ V} \approx 406.15 \text{ V}$

The closest option is 406 V.