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A $100 \mu \mathrm{F}$ capacitor in series with a $40 \Omega$ resistance is connected to a $110 \mathrm{~V}, 60 \mathrm{~Hz}$ supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Solution:
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Verified Answer
Capacitive reactance $X_C=\frac{1}{2 \pi f C}$
$$
X_C=\frac{1}{2 \times \pi \times 60 \times 100 \times 10^{-6}}=26.54 \Omega
$$
Impedence
$$
Z=\sqrt{R^2+X_C^2}=\sqrt{(40)^2+(26.54)^2}=48 \Omega
$$
(a) Virtual current in the circuit
$$
I_v=\frac{E_v}{Z}=\frac{110}{48}=2.29 \mathrm{~A}
$$
Maximum current $I_0=I_v \sqrt{2}=3.24 \mathrm{~A}$
(b) $\tan \phi=\frac{V_C}{V_R}=\frac{1}{\omega C R}$
Phase lag $\phi=\tan ^{-1}\left(\frac{1}{\omega C R}\right)$
$$
=\tan ^{-1}\left(\frac{26.54}{40}\right)
$$
$\phi=33.56^{\circ}=0.186 \pi$ radian
Time lag $t=\phi / \omega=\frac{0.18 \pi}{2 \pi(60)}=1.5 \mathrm{~ms}$
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