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.
Given: In a series L C R circuit, voltage of variable frequency source is 230 V , the resistance is 40 Ω , inductance is 5 H and capacitance is 80 μ F .
a)
The resonance frequency of the circuit is given as,
ω r = 1 L C
Where, the inductance of the given circuit is L and its capacitance is C .
By substituting the values in the above equation, we get
ω r = 1 5 × 80 × 10 − 6 = 50 rad/s
Thus, the source frequency which drives the circuit in resonance is 50 rad/s .
b)
At resonant frequency, the circuit behaves as a purely resistive circuit.
Hence, the impedance of the circuit at resonant frequency is 40 Ω .
Amplitude of current in the circuit is given as,
i m = 2 V Z
By substituting the values in the above equation, we get
i m = 2 × 230 40 = 8.13 A
Thus, amplitude of current in the circuit at resonant frequency is 8.13 A .
c)
At resonant frequency, the rms current in the circuit is given as,
I = V Z
By substituting the values in the above equation, we get
I = 230 40 = 5.75 A
The rms potential drop across the resistive element is given as,
V R = I R
By substituting the values in the above equation, we get
V R = 5.75 × 40 = 230 V
The rms potential drop across the inductive element is given as,
V L = ω L I
By substituting the values in the above equation, we get
V L = 50 × 5 × 5.75 = 1437.5 V
The rms potential drop across the capacitive element is given as,
V C = 1 ω C × I
By substituting the values in the above equation, we get
V C = 1 50 × 80 × 10 − 6 × 5.75 = 1437.5 V
We observe that at resonant frequency,
V L = V C
Thus, potential drop across the L C combination is zero at the resonating frequency.
Given: In a series
a)
The resonance frequency of the circuit is given as,
Where, the inductance of the given circuit is
By substituting the values in the above equation, we get
Thus, the source frequency which drives the circuit in resonance is
b)
At resonant frequency, the circuit behaves as a purely resistive circuit.
Hence, the impedance of the circuit at resonant frequency is
Amplitude of current in the circuit is given as,
By substituting the values in the above equation, we get
Thus, amplitude of current in the circuit at resonant frequency is
c)
At resonant frequency, the rms current in the circuit is given as,
By substituting the values in the above equation, we get
The rms potential drop across the resistive element is given as,
By substituting the values in the above equation, we get
The rms potential drop across the inductive element is given as,
By substituting the values in the above equation, we get
The rms potential drop across the capacitive element is given as,
By substituting the values in the above equation, we get
We observe that at resonant frequency,
Thus, potential drop across the
