The voltage source $v_s=10\sqrt{2}\sin(20000\pi t)$ V has an internal resistance of $50~\Omega$. Find the RMS current through the $R=25~\Omega$ resistor (rounded to one decimal place).
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At series resonance, $X_L=X_C$ so reactances cancel. The entire source voltage drops across the resistances, and $I_{\rm rms}=V_{\rm rms}/R_{\text{tot}}$.
Given $\omega=20000\pi~\text{rad/s}$, $L=\dfrac{1}{20\pi}\,\text{mH}$, $C=\dfrac{1}{20\pi}\,\text{mF}$.
\[
X_L=\omega L=\left(20000\pi\right)\!\left(\frac{10^{-3}}{20\pi}\right)=1~\Omega,\qquad
X_C=\frac{1}{\omega C}=\frac{1}{(20000\pi)\left(\frac{10^{-3}}{20\pi}\right)}=1~\Omega.
\]
Thus the net reactance is zero (series resonance). Total series resistance
\[
R_{\text{tot}}=50+25+25=100~\Omega.
\]
Source RMS voltage $=10$ V $\Rightarrow$
\[
I_{\rm rms}=\frac{10}{100}=0.10~\text{A}=100.0~\text{mA}.
\]
Final Answer: 100.0 mA
