The Total solution for NCERT class 6-12
Figure 7.21 shows a series LCR circuitconnected to a variable frequency 230 V source. L = 5.0H, C = 80μF, R = 40 Ω
(a) Determinethe source frequency which drives the circuit in resonance.
(b) Obtainthe impedance of the circuit and the amplitude of current at the resonatingfrequency.
(c) Determinethe rms potential drops across the three elements of the circuit. Show that thepotential drop across the LC combination is zero at theresonating frequency.
Inductance of the inductor, L =5.0 H
Capacitance of thecapacitor, C = 80 μF = 80 × 10−6 F
Resistance of the resistor, R =40 Ω
Potential of the variable voltagesource, V = 230 V
Hence, the circuit will come inresonance for a source frequency of 50 rad/s.
At resonance,
Amplitude of the current at the resonating frequencyis given as:
Where,
V0 =Peak voltage
Hence, at resonance, theimpedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.
(c) Rmspotential drop across the inductor,
(VL)rms = I × ωRL
I =rms current
Potential drop across thecapacitor,
Potential drop across theresistor,
(VR)rms = IR
=× 40 = 230 V
Potential drop across the LC combination,
∴VLC= 0
Hence, it is proved that thepotential drop across the LC combination is zero at resonatingfrequency.