Question

An $L-C-R$ series circuit with $L=0.120H,R=240\Omega$, and $C=7.30\mu F$ carries an rms current of $0.450A$ with a frequency of $400Hz$. What is the rms voltage of the source?

Answer 1

Given, inductance, $L=0.120H$; capacitance, $C=7.30\times {10}^{-6}f$

resistance, $R=240\Omega$; rms current, $I_{rms}=0.450A$

frequency, $f=400H_z$

Now, the impedance due to inductor, $X_L$ is:

$X_L=2\pi fL=2\pi \left(400\right)\left(0.12\right)$

$X_L=301.6\Omega$

the impedance due to capacitor, $X_c$ is:

$X_c=\frac{1}{2\pi fC}=\frac{1}{2\pi \left(400\right)\left(7.3\right)\times {10}^{-6}}$

$X_c=54.5\Omega$

the total impedance of the LCR circuit, Z is:

$Z=\sqrt{R^2+(X_L-X_c{)}^2}$

$Z=\sqrt{(240{)}^2+(\left(301.6\right)-\left(54.5\right){)}^2}$

or $Z=344\Omega$

Now, we know

$V_{rms}=I_{rms}Z$

$=\left(0.450\right)\left(344\right)$

$=154.8v$

Hene, we get $V_{rms}=155V$