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Question: Answered & Verified by Expert
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?
PhysicsAlternating Current
Solution:
2234 Upvotes 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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