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प्रश्न
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?
उत्तर
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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