An inductor 200 mH, a capacitor 500 μF and a resistor 10 Ω are connected in series with a 100 V variable frequency a.c. source. Calculate the (i) frequency at which the power factor of the circuit is unity. (ii) current amplitude at this frequency. (iii) Q-factor.
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It is given that:
Inductance of the inductor, L = 200 mH = 200 × 10−3 = 0.2 H
Capacitance of the capacitor, C = 500 μF = 500 × 10−6 = 5 × 10−4 F Resistance of the resistor, R = 10 Ω
Voltage of the source, V = 100 V
Power factor of the LCR circuit, cosΦ = 1
(i) Power factor of the LCR circuit is given as: Hence, the power factor of the series LCR circuit will be unity at frequency 16 Hz.
(ii) Current amplitude at resonance frequency is calculated as: This is the rms value of current. Current amplitude is the peak current which can be calculated by multiplying the rms value.
See attachment
Inductance of the inductor, L = 200 mH = 200 × 10−3 = 0.2 H
Capacitance of the capacitor, C = 500 μF = 500 × 10−6 = 5 × 10−4 F Resistance of the resistor, R = 10 Ω
Voltage of the source, V = 100 V
Power factor of the LCR circuit, cosΦ = 1
(i) Power factor of the LCR circuit is given as: Hence, the power factor of the series LCR circuit will be unity at frequency 16 Hz.
(ii) Current amplitude at resonance frequency is calculated as: This is the rms value of current. Current amplitude is the peak current which can be calculated by multiplying the rms value.
See attachment
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