Question

Figure shows a series $LCR$ circuit with $R = 200 \,\Omega,\, C = 15.0\, \mu F$ and $L = 230\, mH$. If $\varepsilon = 36.0\, sin\, 120\pi t$, 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 $R=200\, \Omega$,
$C=15.0\, \mu F=15 \times 10^{-6} F$
$L=230\, mH =230 \times 10^{-3} H$,
$\varepsilon=36.0 \sin 120 \pi t$
Compare it with standard equation
$\varepsilon=\left(\varepsilon_{0} \sin \omega t\right)$
We get
$\varepsilon_{0}=36.0, \omega=120 \pi$
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
$Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}$
$=\sqrt{(200)^{2}+(177-87)^{2}}=219\, \Omega$
$\therefore I_{0}=\frac{\varepsilon_{0}}{Z}$
$=\frac{36 V }{219 \,\Omega}=164\, mA$