Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 W. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
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(a) resonant frequency,
given, L = 5H, C = 80μF = 80 × 10^-6 C
so, resonant frequency = 1/√(5 × 80 × 10^-6)
= 1/√(400 × 10^-6)
= 1/√(0.0004) = 1/0.02 = 50 rad/s
(b) at resonance,
but we know, impedance , Z =
then, impedance, Z = R = 40Ω
now, RMS value of current, = /R
given, emf of source, E = 230V and R = 40Ω
so, = 230/40 = 5.75A
then, amplitude of current, I = √2 = 8.13A
(c) potential drop across ' L'
= 5.75A × (L)
= 5.75A × 50rad/s × 5H
= 1437.5V
potential drop across 'C'
= 5.75 ×
= 5.75A × 1/(50rad/s × 80 × 10^-6C)
= 14.37.5V
potential drop across R,
= 5.75 A × 40Ω = 230V
as and,
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