Question

Figure shows a series $LCR$ circuit with $R=200\Omega ,C=15.0\mu F$ and $L=230mH$. If $\epsilon =36.0sin120\pi t$, the amplitude $I_0$ of the current $I$ in the circuit is close to

• A. 109 mA
• B. 126 mA
• C. 150 mA
• D. 164 mA

Answer 1

Answer: (D)

Given $R=200\Omega$,
$C=15.0\mu F=15\times 10^{-6}F$
$L=230mH=230\times 10^{-3}H$,
$\epsilon =36.0\sin 120\pi t$
Compare it with standard equation
$\epsilon =\left({\epsilon }_0\sin \omega t\right)$
We get
${\epsilon }_0=36.0,\omega =120\pi$
The capacitive reactance is
$X_C=\frac{1}{\omega C}$
$=\frac{1}{120\pi \times 15\times 10^{-6}}=177\Omega$
The inductive reactance is
$X_L=\omega L$
$=120\pi \times 230\times 10^{-3}=87\Omega$
Impedance of the series $LCR$ circuit is
$Z=\sqrt{R^2+{\left(X_C-X_L\right)}^2}$
$=\sqrt{(200{)}^2+(177-87{)}^2}=219\Omega$
$\therefore I_0=\frac{{\epsilon }_0}{Z}$
$=\frac{36V}{219\Omega }=164mA$