1. Modeling the Circuit

The circuit consists of a source $V = V_0 \sin(2\pi f t)$.

• Branch 1 (Series): An inductor $L$ and Voltmeter A (which acts as a resistance $R$) are in series.
• Branch 2 (Parallel Load): The second inductor $L$ is in parallel with Voltmeter B (resistance $R$).

We want to find the maximum reading of Voltmeter B as the frequency changes. Voltmeter B measures the voltage across the parallel combination ($V_p$).

2. Impedance Calculation

Let $X = \omega L$.
Impedance of parallel part ($R$ || $L$): $$ Z_p = \frac{jXR}{R+jX} $$ Total Impedance of the circuit (assuming Series setup based on diagram logic): $$ Z_{total} = (R + jX) + Z_p = (R + jX) + \frac{jXR}{R+jX} $$ The voltage across Voltmeter B is determined by the voltage divider rule: $$ V_B = V_0 \left| \frac{Z_p}{Z_{total}} \right| $$ Substituting and simplifying: $$ \frac{Z_p}{Z_{total}} = \frac{\frac{jXR}{R+jX}}{R+jX + \frac{jXR}{R+jX}} = \frac{jXR}{(R+jX)^2 + jXR} $$ $$ = \frac{jXR}{R^2 – X^2 + 2jRX + jXR} = \frac{jXR}{R^2 – X^2 + 3jRX} $$

3. Optimization

The magnitude of the voltage is: $$ |V_B| = V_0 \frac{XR}{\sqrt{(R^2-X^2)^2 + (3RX)^2}} $$ To maximize this, we need to minimize the denominator term relative to the numerator. Dividing numerator and denominator by $XR$: $$ |V_B| = V_0 \frac{1}{\sqrt{\frac{(R^2-X^2)^2}{X^2 R^2} + 9}} = V_0 \frac{1}{\sqrt{\left(\frac{R}{X} – \frac{X}{R}\right)^2 + 9}} $$ The term $\left(\frac{R}{X} – \frac{X}{R}\right)^2$ is minimized when $\frac{R}{X} = \frac{X}{R}$, i.e., when $X = R$.

4. Maximum Value

At $X = R$, the square term becomes zero.

$$ |V_B|_{max} = V_0 \frac{1}{\sqrt{0 + 9}} = \frac{V_0}{3} $$