Question

An AC generator producing $10V$ (rms) at $200rad/s$ is connected in series with a $50\Omega$ resistor, a $400mH$ inductor and a $200\mu F$ capacitor. The rms voltage across the inductor is

• A. $2.5V$
• B. $3.4V$
• C. $6.7V$
• D. $10.8V$

Answer 1

The correct option is D $10.8V$

Given : $E=10V,w=200,R=50\Omega ,L=400mH,C=200\mu F$
So, capacitive reactance $X_C=\frac{1}{wC}=\frac{1}{\left(200\right)(200\times {10}^{-6})}=25\Omega$
Inductive reactance $X_L=wL=\left(200\right)(400\times {10}^{-3})=80\Omega$
Impedance $Z=\sqrt{R^2+{(X_L-X_C)}^2}=\sqrt{{50}^2+{(80-25)}^2}$
$Z=74.3\Omega$
$I=\frac{E}{Z}=\frac{10}{74.3}=0.13459A$
$E_L=I{X}_L=0.1345\times 80=10.76V$