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Home » A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 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 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 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?
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A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 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?

Capacitance of the capacitor, \mathrm{C}=100 \mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}

The resistance of the resistor =40 \Omega

Supply voltage, \mathrm{V}=110 \mathrm{~V}

Frequency, v=60 \mathrm{~Hz}

Angular frequency, \omega=2 \pi \mathrm{V}=2 \pi \times 60=120 \pi \mathrm{rad} / \mathrm{s}

(a) Maximum current in the circuit

    \[I_{0}=V_{0} / Z\]

Here, Z=\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}

Z=\sqrt{40^{2}+\frac{1}{(120 \pi)^{2}\left(10^{-4}\right)^{2}}}

Z=\sqrt{1600+\frac{10^{8}}{(120 \pi)^{2}}}

Z=\sqrt{1600+\frac{10^{8}}{141978}}

Z=\sqrt{1600+704}

Z=\sqrt{2304}

\text{Z}=48

{{\text{I}}_{0}}=110\sqrt{2/48}

{{\text{I}}_{0}}=155.6/48=3.24

(b) In the capacitor circuit, the voltage will lag behind the current by a phase angle \Phi.

Therefore, \tan \Phi=1 / \omega \mathrm{CR}

=1/120\pi \times {{10}^{-4}}\times 40

=0.6635

\Phi ={{\tan }^{-1}}(0.6635)={{33.56}^{0}}

=33.56\pi /180\text{rad}

Time lag, \mathrm{t}=\Phi / \omega=33.56 \pi /(180 \times 120 \pi)=1.55 \times 10^{-3} \mathrm{~S}=1.55 \mathrm{~ms}

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