Class 12 Physics
CBSE and UP Board
Alternating Current
Chapter-7 Exercise 7.16
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 7 Question 16
Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
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Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F =10⁻⁴F
Resistance of the resistor, R = 40Ω
Supply voltage, V = 110 V
Frequency of the supply, ν = 12 kHz = 12 x 103 Hz
Angular Frequency, ω = 2π ν = 2 x π x 12 x 103 Hz
= 24π x 103 rad/s
Peak voltage, V0 = V√2 = 110√2 V
Maximum Current ,I0= V0/√(R2 +1/ω2c2)
=110 √2 /√[(40)2 +1/(24π x 103)2(10⁻⁴)2]
=110 √2/√[1600 + (10/24π)2] = 3.9 A
For an RC circuit, the voltage lags behind the current by a phase angle of φ given as:
tan φ = (1/ωC)/R = 1/ωCR = 1/(24π x 103 x 10⁻⁴ x 40)
tan φ = 1/96π
Therefore , φ = 0.2º or 0.2 π/180 rad
Therefore ,Time lag = φ/ω = 0.2 π/(180 x 24π x 103)
=4.69 x 10⁻³s
= 0.04μs
Hence, φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor. In a dc circuit, after the steady state is achieved, ω = 0.
Hence, capacitor C amounts to an open circuit.