Physics, asked by Draj3283, 11 months ago

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.

Answers

Answered by abhi178
30

(a) resonant frequency, \omega_r=\frac{1}{\sqrt{LC}}

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, X_C=X_L

but we know, impedance , Z = \sqrt{R^2+(X_C-X_L)^2}

then, impedance, Z = R = 40Ω

now, RMS value of current, I_{rms} = E_{rms}/R

given, emf of source, E = 230V and R = 40Ω

so, I_{rms} = 230/40 = 5.75A

then, amplitude of current, I = √2I_R = 8.13A

(c) potential drop across ' L'

V_L=I_{rms}X_L

= 5.75A × (\omega_rL)

= 5.75A × 50rad/s × 5H

= 1437.5V

potential drop across 'C'

V_C=I_{rms}X_C

= 5.75 × \frac{1}{\omega C}

= 5.75A × 1/(50rad/s × 80 × 10^-6C)

= 14.37.5V

potential drop across R,

V_R=I_{rms}R

= 5.75 A × 40Ω = 230V

as V_L-V_C=0 and, E_{rms}=V_R=230V

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