Two identical Inductor

Let the resistance of the voltmeters be $R$ , the inductance of the inductors be $L$ , and $2\pi f = \omega$ . Then the reading on voltmeter $B$ is:

$\begin{aligned} V_B & = \frac {\frac {j\omega RL}{R+j\omega L}}{R+j\omega L + \frac {j\omega RL}{R+j\omega L}} = \frac {j\omega RL}{(R+j\omega L)^2 + j\omega RL} = \frac {j\omega RL}{R^2 - \omega^2 L^2 + 3j\omega RL} = \frac 1{3+j\frac {\omega^2L^2 - R^2}{\omega RL}} \end{aligned}$

$\begin{aligned} \implies |V_B| & = \frac 1{\left|3+j\frac {\omega^2L^2 - R^2}{\omega RL}\right|} \\ & = \frac 1{\sqrt{3^2 + \left(\frac {\omega^2L^2 - R^2}{\omega RL}\right)^2}} \\ & = \frac 1{\sqrt{\blue{\frac {\omega^2L^2}{R^2}}+7+\blue{\frac {R^2}{\omega^2L^2}}}} & \small \blue{\text{By AM-GM inequality: }\frac {\omega^2L^2}{R^2}+\frac {R^2}{\omega^2L^2} \ge 2} \\ & \le \frac 1{\sqrt {\blue 2+7}} = \frac 13 \approx \boxed{0.333} & \small \blue{\text{Equality occurs when }R=\omega L} \end{aligned}$

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Reference: AM-GM inequality

Let the resistance of the voltmeter be $1$ . The choice of resistance value does not affect the max value of $V_B$ , even though it does affect the value of inductive reactance at which it occurs.

Define the series and parallel $RL$ impedances as follows:

$Z_S = 1 + jX \\ Z_P = \frac{j X}{1 + j X}$

Then the voltage across voltmeter $B$ is:

$V_B = V_0 \frac{Z_P}{Z_S + Z_P} = V_0 \frac{j X}{(1 + jX)^2 + jX}$

The magnitude of this quantity is:

$|V_B| = V_0 \frac{X}{\sqrt{1 + 7X^2 + X^4}}$

Differentiate the above expression with respect to $X$ and set it to zero to find the max value. The maximum value of $V_B$ is $\frac{1}{3} V_0$ when $X = 1$ .