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Question
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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Solution
Capacitance of the capacitor, C = 100 μF = 100 × 10−6 F
Resistance of the resistor, R = 40 Ω
Supply voltage, V = 110 V
(a) Frequency of oscillations, v = 60 Hz
Angular frequency, ω = 2πv = 2π × 60 rad/s
For an RC circuit, we have the relation for impedance as:
Z = `"R"^2 + 1/(ω^2"C"^2)`
Peak voltage, V0 = `"V"sqrt2 = 110 sqrt2 "V"`
Maximum current is given as:
I0 = `"V"_0/"Z"`
= `"V"_0/sqrt("R"^2 + 1/(ω^2"C"^2))`
= `(110 sqrt2)/sqrt((40)^2 + 1/((120π)^2 xx (10^-4)^2))`
= `(110 sqrt2)/sqrt(1600 + 10^8/(120π)^2)`
= 3.24 A
(b) In a capacitor circuit, the voltage lags behind the current by a phase angle of Φ. This angle is given by the relation:
∴ `tan phi = (1/(ω"C"))/"R" = 1/(ω"CR")`
= `1/(120π xx 10^-4 xx 10)`
= 0.6635
`phi` = tan−1 (0.6635) = 33.56°
= `(33.56π)/180 "rad"`
∴ Time lag = `phi/ω`
= `(33.56π)/(180 xx 120π)`
= 1.55 × 10−3 s
= 1.55 ms
Hence, the time lag between maximum current and maximum voltage is 1.55 ms.
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