At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit. Resistance, R = 20 Ω Inductance, L = 1.5 H Capacitance, C = 35 μF = 30 x 10_6F AC supply voltage to the LCR circuit, V = 200 V Impedance of the circuit is given by the relation, Z = √ [R2 + (xL-xc)Read more
At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.
Resistance, R = 20 Ω
Inductance, L = 1.5 H
Capacitance, C = 35 μF = 30 x 10_6F
AC supply voltage to the LCR circuit, V = 200 V
Impedance of the circuit is given by the relation,
Z = √ [R2 + (xL-xc)2 ]
At resonance, XL = Xc :.Z = R = 20 Ω
Current in the circuit can be calculated as:
I = V/Z = 200/20= 10 A
Hence, the average power transferred to the circuit in one complete cycle:
Capacitance of the capacitor, C = 30μF = 30 x 10⁻6F Inductance of the inductor, L = 27 mH = 27 x 10⁻3 H Charge on the capacitor, Q = 6 mC = 6 x 10⁻3 C Total energy stored in the capacitor can be calculated as: E = 1/2 x Q2/C = 1/2 x (6 x 10⁻3)2/(30 x 10⁻6) = 6/10 = 0.6 J Total energy at a later tRead more
Capacitance of the capacitor, C = 30μF = 30 x 10⁻6F
Inductance of the inductor, L = 27 mH = 27 x 10⁻3 H
Charge on the capacitor, Q = 6 mC = 6 x 10⁻3 C
Total energy stored in the capacitor can be calculated as:
E = 1/2 x Q2/C = 1/2 x (6 x 10⁻3)2/(30 x 10⁻6) = 6/10 = 0.6 J
Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.
Capacitance, C = 30μF = 30 x 10⁻6F Inductance, L = 27 mH = 27 x 10-3 H Angular frequency is given as: ωr = 1 /√(LC) = 1/√ ( 27 x 10-3 x 30 x 10⁻6 ) = 1/ (9 x 10⁻⁴ ) rad/s = 1.11 x 10³ rad/s Hence ,the angular frequency of free oscillations of the circuit is 1.11 x 10³ rad/s
Capacitance, C = 30μF = 30 x 10⁻6F
Inductance, L = 27 mH = 27 x 10-3 H
Angular frequency is given as:
ωr = 1 /√(LC) = 1/√ ( 27 x 10-3 x 30 x 10⁻6 )
= 1/ (9 x 10⁻⁴ ) rad/s
= 1.11 x 10³ rad/s
Hence ,the angular frequency of free oscillations of the circuit is 1.11 x 10³ rad/s
Inductance, L = 2.0 H Capacitance, C = 32 μF = 32 x 10-6 F Resistance, R = 10Ω Resonant frequency is given by the relation, ωr= 1/ √ (LC) = 1/ √ (2 x 32 x 10-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s Now ,Q-value of the circuit is given as : Q = 1/R√ (L/C) = (1/10) x √[2/(32 x 10-6 ) ] = 1/(10 x 4 x 10⁻³ ) =Read more
Inductance, L = 2.0 H
Capacitance, C = 32 μF = 32 x 10-6 F
Resistance, R = 10Ω
Resonant frequency is given by the relation,
ωr= 1/ √ (LC) = 1/ √ (2 x 32 x 10-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s
In the inductive circuit, rms value of current, I = 15.92 A rms value of voltage, V = 220 V Hence, the net power absorbed can be obtained by the relation, P = VI cos φ Where, φ = Phase difference between V and I. For a pure inductive circuit, the phase difference between alternating voltage and currRead more
In the inductive circuit,
rms value of current, I = 15.92 A
rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos φ
Where, φ = Phase difference between V and I.
For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit, rms value of current, I = 2.49 A, rms value of voltage, V = 110 V Hence, the net power absorbed can be obtained as:
P = VI Cos φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e.,
A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit. Resistance, R = 20 Ω Inductance, L = 1.5 H Capacitance, C = 35 μF = 30 x 10_6F AC supply voltage to the LCR circuit, V = 200 V Impedance of the circuit is given by the relation, Z = √ [R2 + (xL-xc)Read more
At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.
Resistance, R = 20 Ω
Inductance, L = 1.5 H
Capacitance, C = 35 μF = 30 x 10_6F
AC supply voltage to the LCR circuit, V = 200 V
Impedance of the circuit is given by the relation,
Z = √ [R2 + (xL-xc)2 ]
At resonance, XL = Xc
:.Z = R = 20 Ω
Current in the circuit can be calculated as:
I = V/Z = 200/20= 10 A
Hence, the average power transferred to the circuit in one complete cycle:
VI = 200 x 10 = 2000 W.
See lessSuppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Capacitance of the capacitor, C = 30μF = 30 x 10⁻6F Inductance of the inductor, L = 27 mH = 27 x 10⁻3 H Charge on the capacitor, Q = 6 mC = 6 x 10⁻3 C Total energy stored in the capacitor can be calculated as: E = 1/2 x Q2/C = 1/2 x (6 x 10⁻3)2/(30 x 10⁻6) = 6/10 = 0.6 J Total energy at a later tRead more
Capacitance of the capacitor, C = 30μF = 30 x 10⁻6F
Inductance of the inductor, L = 27 mH = 27 x 10⁻3 H
Charge on the capacitor, Q = 6 mC = 6 x 10⁻3 C
Total energy stored in the capacitor can be calculated as:
E = 1/2 x Q2/C = 1/2 x (6 x 10⁻3)2/(30 x 10⁻6) = 6/10 = 0.6 J
Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.
See lessA charged 30μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Capacitance, C = 30μF = 30 x 10⁻6F Inductance, L = 27 mH = 27 x 10-3 H Angular frequency is given as: ωr = 1 /√(LC) = 1/√ ( 27 x 10-3 x 30 x 10⁻6 ) = 1/ (9 x 10⁻⁴ ) rad/s = 1.11 x 10³ rad/s Hence ,the angular frequency of free oscillations of the circuit is 1.11 x 10³ rad/s
Capacitance, C = 30μF = 30 x 10⁻6F
Inductance, L = 27 mH = 27 x 10-3 H
Angular frequency is given as:
ωr = 1 /√(LC) = 1/√ ( 27 x 10-3 x 30 x 10⁻6 )
= 1/ (9 x 10⁻⁴ ) rad/s
= 1.11 x 10³ rad/s
Hence ,the angular frequency of free oscillations of the circuit is 1.11 x 10³ rad/s
See lessObtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32μF and R = 10 Ω. What is the Q-value of this circuit?
Inductance, L = 2.0 H Capacitance, C = 32 μF = 32 x 10-6 F Resistance, R = 10Ω Resonant frequency is given by the relation, ωr= 1/ √ (LC) = 1/ √ (2 x 32 x 10-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s Now ,Q-value of the circuit is given as : Q = 1/R√ (L/C) = (1/10) x √[2/(32 x 10-6 ) ] = 1/(10 x 4 x 10⁻³ ) =Read more
Inductance, L = 2.0 H
Capacitance, C = 32 μF = 32 x 10-6 F
Resistance, R = 10Ω
Resonant frequency is given by the relation,
ωr= 1/ √ (LC) = 1/ √ (2 x 32 x 10-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s
Now ,Q-value of the circuit is given as :
Q = 1/R√ (L/C) = (1/10) x √[2/(32 x 10-6 ) ]
= 1/(10 x 4 x 10⁻³ ) = 25
Hence, the Q-Value of this circuit is 25.
See lessIn Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
In the inductive circuit, rms value of current, I = 15.92 A rms value of voltage, V = 220 V Hence, the net power absorbed can be obtained by the relation, P = VI cos φ Where, φ = Phase difference between V and I. For a pure inductive circuit, the phase difference between alternating voltage and currRead more
In the inductive circuit,
rms value of current, I = 15.92 A
rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos φ
Where, φ = Phase difference between V and I.
For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit, rms value of current, I = 2.49 A, rms value of voltage, V = 110 V Hence, the net power absorbed can be obtained as:
P = VI Cos φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e.,
φ = 90°.
Hence, P = 0 i.e., the net power is zero.
See less