Question

In a series LCR circuit with an AC source, $R=300\Omega ,C=20\mu F,L=10henry{ϵ}_{max}=50V$ and $v=50/\pi Hz$. Find
(a) the $rms$ current in the circuit.
(b) the $rms$ potential differences across the capacitor, the resistor and the inductor, Note that the sum of the $rms$ potential difference across the three elements is greater than the $rms$ voltage of the source.

Answer 1

Given: $R=300\Omega$,$C=20\mu F=20\times {10}^{-6}$F and $L=1$Henry, $E=50$V, $V=\frac{50}{\pi }$Hz.

$\left(a\right)I_0=\frac{E_0}{Z}$

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

$=\sqrt{{300}^2+{(\frac{1}{2\pi fC}-2\pi fL)}^2}$

$=\sqrt{{300}^2+{(\frac{1}{2\pi \times \frac{50}{\pi }\times 20\times {10}^{-6}}-2\pi \times \frac{50}{\pi }\times 1)}^2}$

$=\sqrt{{300}^2+{(\frac{{10}^4}{20}-100)}^2}=500$

$I_0=\frac{E_0}{Z}=\frac{50}{500}=0.1$A

$\left(b\right)$Potential across the capacitor$=i_0\times X_c=0.1\times 500=50$V

Potential difference across the resistor $i_0\times R=0.1\times 300=30V$

Potential difference across the inductor $i_0\times X_L=0.1\times 100=10V$

Rms.Potential$=50$V

Net sum of all potential drops$=50$V$+30$V$+10$V$=90$V

Sum of potential drops$>$R.M.S potential applied.