Answer
Please see the work below.
Work Step by Step
(a) We can find the required voltmeter reading between the points A and B as follows:
$\omega=2\pi f$
$\omega=2\pi(30,000)=60,000\pi s^{-1}$
The inductive reactance $X_L=\omega L$
$X_L=60,000\pi\times 0.3\times 10^{-3}=56.57\Omega$
The capacitive reactance is $X_C=\frac{1}{\omega C}$
$X_C=\frac{1}{60,000\pi\times 0.1\times 10^{-6}}$
$X_C=53\Omega$
The impedance of the circuit is given as $Z=\sqrt{R^2+(X_L-X_C)^2}$
$Z=\sqrt{(2.5)^2+(56.57-53)^2}$
$\implies Z=4.36\Omega$
The rms current $I_{rms}=\frac{V_{rms}}{Z}$
$\implies I_{rms}=\frac{6}{4.36}=1.4A$
Now the voltmeter reading between the points A and B is
$V_{rms,L}=I_{rms}\times X_L$
We plug in the known values to obtain:
$V_{rms,L}=1.4\times 56.57=79.198V$
(b) The voltmeter reading between the points B and C is the rms voltage across the capacitor
$V_{rms,C}=I_{rms}\times X_C$
We plug in the known values to obtain:
$V_{rms,C}=1.4\times 53=74.2V$
(c) The voltmeter reading between the points A and C is the rms voltage across the inductor and capacitor.
$V_{rms,LC}=I_{rms}\times(X_L-X_C)$
We plug in the known values to obtain:
$V_{rms,LC}=1.4(56.57-53)=4.998V$
(d) The voltmeter reading between the points A and D is the voltage across RLC. This is actually the rms voltage in the circuit that is $V_{rms,RLC}=V_{rms}=6V$
