Inductance of the inductor, L = 5.0 H, Capacitance of the capacitor, C = 80 μH = 80 x 10-6 F Resistance of the resistor, R = 40Ω Potential of the variable voltage source, V = 230 V Ans (a). Resonance angular frequency is given as: ωr = 1/√(LC) = 1/ √(5 x 80 x 10-6) = 10³/20 = 50 rad/s Hence, the cirRead more
Inductance of the inductor, L = 5.0 H,
Capacitance of the capacitor, C = 80 μH = 80 x 10-6 F
Resistance of the resistor, R = 40Ω
Potential of the variable voltage source, V = 230 V
Ans (a).
Resonance angular frequency is given as:
ωr = 1/√(LC) = 1/ √(5 x 80 x 10-6) = 10³/20 = 50 rad/s
Hence, the circuit will come in resonance for a source frequency of 50 rad/s.
Ans (b).
Impedance of the circuit is given by the relation:
Z = √ [R² +( XL –XC)² ]
At resonance, XL = Xc => Z = R = 40Ω
Amplitude of the current at the resonating frequency is given as: Io = Vo/Z
Where, V0 = Peak voltage = √2V
Therefore, Io =√2V/Z = √2 x 230/40 = 8.13 A
Hence, at resonance, the impedance of the circuit is 40Ω and the amplitude of the current is 8.13 A.
Ans (c).
rms potential drop across the inductor,
(VL)rms = I x ωrL
Where,
Irms = Io/√2 = √2V/√2Z= 230/40 = 23/4 A
Therefore
(VL)rms = (23/4) x 50 x 5 =1437.5 V
Potential drop across the capacitor:
(VC)rms = I x 1/ωrC = (23/4 ) x 1/(50 x 80 x 10-6) =1437.5 V
Potential drop across the resistor:
(VR)rms =IR = (23/4) x 40 = 230V
Potential drop across the LC combination:
VLC = I (XL-Xc)
At resonance, XL = Xc => VLC = 0
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.
The range of frequency (ν) of a radio is 800 kHz to 1200 kHz. Lower tuning frequency, ν1 = 800 kHz = 800 x 103 Hz Upper tuning frequency, ν2 = 1200 kHz = 1200 x 103 Hz Effective inductance of circuit L = 200μH = 200 x 10⁻6 H Capacitance of variable capacitor for ν1 is given as: C₁ = 1/(ω1)²L Where,Read more
The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.
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
Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at e ^ 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.
Inductance of the inductor, L = 5.0 H, Capacitance of the capacitor, C = 80 μH = 80 x 10-6 F Resistance of the resistor, R = 40Ω Potential of the variable voltage source, V = 230 V Ans (a). Resonance angular frequency is given as: ωr = 1/√(LC) = 1/ √(5 x 80 x 10-6) = 10³/20 = 50 rad/s Hence, the cirRead more
Inductance of the inductor, L = 5.0 H,
Capacitance of the capacitor, C = 80 μH = 80 x 10-6 F
Resistance of the resistor, R = 40Ω
Potential of the variable voltage source, V = 230 V
Ans (a).
Resonance angular frequency is given as:
ωr = 1/√(LC) = 1/ √(5 x 80 x 10-6) = 10³/20 = 50 rad/s
Hence, the circuit will come in resonance for a source frequency of 50 rad/s.
Ans (b).
Impedance of the circuit is given by the relation:
Z = √ [R² +( XL –XC)² ]
At resonance, XL = Xc => Z = R = 40Ω
Amplitude of the current at the resonating frequency is given as: Io = Vo/Z
Where, V0 = Peak voltage = √2V
Therefore, Io =√2V/Z = √2 x 230/40 = 8.13 A
Hence, at resonance, the impedance of the circuit is 40Ω and the amplitude of the current is 8.13 A.
Ans (c).
rms potential drop across the inductor,
(VL)rms = I x ωrL
Where,
Irms = Io/√2 = √2V/√2Z= 230/40 = 23/4 A
Therefore
(VL)rms = (23/4) x 50 x 5 =1437.5 V
Potential drop across the capacitor:
(VC)rms = I x 1/ωrC = (23/4 ) x 1/(50 x 80 x 10-6) =1437.5 V
Potential drop across the resistor:
(VR)rms =IR = (23/4) x 40 = 230V
Potential drop across the LC combination:
VLC = I (XL-Xc)
At resonance, XL = Xc => VLC = 0
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.
See lessA radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200μH, what must be the range of its variable capacitor? [Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.]
The range of frequency (ν) of a radio is 800 kHz to 1200 kHz. Lower tuning frequency, ν1 = 800 kHz = 800 x 103 Hz Upper tuning frequency, ν2 = 1200 kHz = 1200 x 103 Hz Effective inductance of circuit L = 200μH = 200 x 10⁻6 H Capacitance of variable capacitor for ν1 is given as: C₁ = 1/(ω1)²L Where,Read more
The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.
Lower tuning frequency, ν1 = 800 kHz = 800 x 103 Hz
Upper tuning frequency, ν2 = 1200 kHz = 1200 x 103 Hz
Effective inductance of circuit L = 200μH = 200 x 10⁻6 H
Capacitance of variable capacitor for ν1 is given as:
C₁ = 1/(ω1)²L
Where, ω1 = Angular frequency for capacitor C₁= 2πν1= 2π x 800 x 103 rad/s
Therefore,
C₁ = 1/(2π x 800 x 10³ )² x 200 x 10⁻6 = 1.9809 x 10⁻¹⁰ F = 198 pF
Capacitance of variable capacitor for ν2 is given as ;
C2 = 1/(ω2)²L
Where, ω2 = Angular frequency for capacitor C2 = 2πν2 = 2π x 1200 x 103 rad/s
Therefore ,
C2 = 1/(2π x 1200 x 10³ )² x 200 x 10⁻6 = 0.8804 x 10⁻¹⁰ F = 88 pF
Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.
See lessA 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 less