Physics - EMI/AC Question with Solution | TestHub

Given $V=220\sin (2\times 10^{3}t)$ V, the peak voltage is $V_m=220$ V and the angular frequency is $\omega =2\times 10^3$ rad/s.

The inductive reactance is $X_L=\omega L=(2\times 10^3)\times (10\times 10^{-3})=20\Omega$.

The capacitive reactance is $X_C=\frac{1}{\omega C}=\frac{1}{(2\times 10^3)\times (25\times 10^{-6})}=\frac{1}{50\times 10^{-3}}=20\Omega$.

Since $X_L=X_C$, the circuit is in resonance. The impedance of the circuit is $Z=\sqrt{R^2+(X_L-X_C{)}^2}=R=100\Omega$.

The current amplitude is $I_m=\frac{V_m}{Z}=\frac{220}{100}=2.2$ A.

Answer: $2.2$ A