Question

The current lags the emf by $30°$ in a series RLC circuit with $E_0=10V$ and $R=50\Omega$. What is the peak current through the circuit?

Short answer

Peak current through RLC circuit is, $0.173A$

Step-by-step solution

Given information

We are given that,

Peak voltage is, $E_0=10V$

Resistance is, $R=50\Omega$

Phase difference is,$\varphi =30°$

Simplify

Let inductive reactance is,$X_L$

And capacitive reactance is,$X_C$

We know that,

$$\begin{gathered} \tan \left(\varphi \right)=\frac{X_L-X_C}{R} \\ \tan \left(30°\right)=\frac{X_L-X_C}{50} \\ \frac{1}{\sqrt{3}}=\frac{X_L-X_C}{50} \\ {X}_L-X_C=\frac{50}{\sqrt{3}} \\ {\left(X_L-X_C\right)}^2={\left(\frac{50}{\sqrt{3}}\right)}^2 \\ {\left(X_L-X_C\right)}^2=\frac{2500}{3}=833.3\Omega \end{gathered}$$

Therefore, Impedance is,

$$\begin{gathered} Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2} \\ =\sqrt{{50}^2+833.3} \\ =\sqrt{2500+833.3} \\ =57.73\Omega \end{gathered}$$

Hence Peak current is,

localid="1650491817289" $$\begin{gathered} I_0=\frac{E_0}{Z} \\ =\frac{10}{57.73} \\ =0.173A \end{gathered}$$