Question

A $100V$ AC source of frequency $500Hz$ is connected to an LCR circuit with $L=8.1mH$, $C=12.5\mu F$ and $R=10\Omega$, all connected in series. The potential difference across the resistance will be:

Answer 1

Step 1: Given Data

The voltage of the AC source $V_{rms}=100V$

Frequency of the AC source $f=500Hz$

The inductance of the inductor $L=8.1mH=8.1\times {10}^{-3}H$

The capacitance of the capacitor $C=12.5\mu F=12.5\times {10}^{-6}F$

Resistance of the resistor $R=10\Omega$

Step 2: Calculate the reactances

The reactance of the inductor

$X_L=\omega L=2\mathrm{\pi fL}$

$=2\times 3.14\times 500\times 8.1\times {10}^{-3}$

$=25.4\Omega$

The reactance of the capacitor

$X_C=\frac{1}{\omega C}=\frac{1}{2\mathrm{\pi fC}}$

$=\frac{1}{2\times 3.14\times 500\times 12.5\times {10}^{-6}}$

$=25.4\Omega$

We know that the total impedance on an LCR circuit is given as

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

$=\sqrt{{10}^2+{\left(25.4-25.4\right)}^2}$

$=10\Omega$

Step 3: Calculate the required Voltage

We know that,

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

$=\frac{100}{10}$

$=10A$

Therefore, the potential across the resistor,

$V_R=I_{rms}R$

$=10\times 10=100V$

Hence, the potential difference across the resistance will be $100V$.