Answer
Please see the work below.
Work Step by Step
We can find the required rms voltage as follows:
The impedance of the circuit is
$Z=\sqrt{R^+(2\pi fL-\frac{1}{2\pi fC})^2}$
$\implies Z=\sqrt{(9.9\times 10^3\Omega)^2+(2\pi\times 60.0Hz\times 25\times 10^{-3}H-\frac{1}{2\pi(60.0Hz)(0.15\times 10^{-6}F)})^2}$
$Z=2.026\times 10^4\Omega$
Now $I_{rms}=\frac{V_{rms}}{Z}$
$I_{rms}=\frac{115V}{2.026\times 10^4\Omega}$
$I_{rms}=5.676\times 10^{-3}A$
The rms voltage across the inductor is can be determined as
$V_{rms,L}=I_{rms}\times X_L$
$V_{rms,L}=I_{rms}(2\pi fL)$
$V_{rms,L}=2\pi(5.676\times 10^{-3}A)(60.0Hz)(25\times 10^3H)=53mV$
The rms voltage across the capacitor is
$V_{rms,C}=I_{rms}(\frac{1}{2\pi fC})$
$V_{rms,C}=(5.676\times 10^{-3}A)(\frac{1}{2\pi(60.0Hz)(0.15\times 10^{-6}F)})=100V$
Now the rms voltage across the resistor is
$V_{rms,R}=I_{rms}R$
We plug in the known values to obtain:
$V_{rms,R}=(5.676\times 10^{-3}A)(9.9\times 10^3\Omega)=56V$
