Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40Ω.
(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.
(a) Given: L = 5.0 H
C = 80 Μf = 80 × 10-6F
R = 40Ω
Voltage source V = 230 V
The diagram is given as:
Resonance angular frequency can be calculated as follows
ωR = 1/√LC
ωR = 1/ √(5H × 80 × 10-6F)
On calculating, we get
ωR = 50 rad/s
Therefore, source frequency which drives the circuit in resonance frequency ωR = 50 rad/s.
(b) Impedance can be calculated by using the formula:
At resonance condition i.e. when 
Z = R = 40 Ω
At resonating frequency, the amplitude of the current can be calculated using the formula:
Where V0 is the peak voltage
Z is the impedance
V0 = √2Vrms
I0 = (√2 × 230V)/40Ω
⇒ I0 = 8.13 A
The impedance of the circuit is 40 Ω and the amplitude of current at the resonating frequency is 8.13 A.
(c) The root mean square potential drop across the inductor is given by the following formula:
(VL) rms = I × 1/ωR L
Where I = I0/√ 2
I = √ 2V/√2Z
Substituting values, we get
I = 230V/40Ω = 230/40 A
Now,
(VL) rms = I × 1/ωR L
(VL)r.m.s.
(VL)r.m.s = 1437.5 V
Potential drop across capacitor can be calculated as follows:
(VC) rms = I × 1/ω R C
(VC)r.m.s.
(VC)r.m.s = 1437.5 V
Potential drop across resistor can be calculated as follows:
(VR) rms = I × R
(VR) r. m. s
(VR) rms = 230 V
Potential drop across LC combination can be calculated as follows:
VLC 
At resonance condition i.e. when ωL = 1/ωC
Therefore, VLC = 0