Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F = 10⁻⁴ F Resistance of the resistor, R = 40 Ω Supply voltage, V = 110 V Ans (a). Frequency of oscillations, ν= 60 Hz Angular frequency, ω = 2π ν = 2π x 60 rad/s =120π rad/s For a RC circuit, we have the relation for impedance as: Z = √(R2 +1/ω2Read more
Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F = 10⁻⁴ F
Inductance of the inductor, L = 0.5 Hz, Resistance of the resistor, R = 100 Ω Potential of the supply voltages, V = 240 V, Frequency of the supply,ν = 10 kHz = 104 Hz Angular frequency, ω = 2πν= 2π x 104 rad/s Ans (a). Peak voltage, V0Read more
Inductance of the inductor, L = 0.5 Hz, Resistance of the resistor, R = 100 Ω
Potential of the supply voltages, V = 240 V,
Frequency of the supply,ν = 10 kHz = 104 Hz
Angular frequency, ω = 2πν= 2π x 104 rad/s
Ans (a).
Peak voltage, V0 = √2 V= 240 √2 V
Maximum Current, I0 = V0 /√ (R² + ω² L²)
=240 √2/√[(100)² + (2π x 104 )² x (0.50)² ]
=1.1 x 10⁻²A
Ans (b).
For phase difference φ,we have the relation :
tanφ = Lω/R =(2π x 104 x 0.5)/100 = 100π
=> φ = 89.82º =89.82π /180 rad
ωt = 89.82π /180
=> t = 89.82π /(180 x 2π x 104) = 25μs It can be observed that Io is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit. In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.
Inductance of the inductor, L = 20 mH = 20 x 10⁻3 H Capacitance of the capacitor, C = 50 pF = 50 x 10⁻6 F Initial charge on the capacitor, Q = 10 mC = 10 x 10-3 C Ans (a). Total energy stored initially in the circuit is given as: E = 1/2 x Q²/C = (10 x 10-3)²/ (2 x 50 x 10⁻6 ) = 1 J Hence, the totaRead more
Inductance of the inductor, L = 20 mH = 20 x 10⁻3 H
Capacitance of the capacitor, C = 50 pF = 50 x 10⁻6 F
Initial charge on the capacitor, Q = 10 mC = 10 x 10-3 C
Ans (a).
Total energy stored initially in the circuit is given as:
E = 1/2 x Q²/C = (10 x 10-3)²/ (2 x 50 x 10⁻6 ) = 1 J
Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.
Ans (b).
Natural frequency of the circuit is given by the relation,
ν = 1/[2π√(LC)] = 1/(2π√(20 x 10⁻3 x 50 x 10⁻6) = 10³/2π = 159.24 Hz
Natural angular frequency, ωr = 1/√(LC)
= 1/ √(20 x 10⁻3 x 50 x 10⁻6) = 1/ √ 10⁻6= 103 rad/s
Hence, the natural frequency of the circuit is 103 rad/s.
Ans (c).
(i) For time period (T = 1/ν =1/159.24 = 6.28 ms), total charge on the capacitor at time t,
Q’ = Qcos2πt/T
For energy stored is electrical, we can write Q’ = Q.
Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t =0,T/2,T,3T/2
(ii) Magnetic energy is the maximum when electrical energy, Q’ is equal to 0.
Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time, t = T/4 ,3T/4,5T/4—-
Ans (d).
Q1 = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.
When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = 1/2(maximum energy)
=>1/2 x (Q¹)² /C =1/2 x (1/2C x Q²) = Q²/4C
Q¹ =Q/√2
But Q¹ = Q cos(2π/T) .t
=> Q/√2= Q cos(2π/T) .t
=> cos(2π/T) .t= 1/√2 = cos (2n+1)π/4; where n= 0,1,2,…
t= (2n+1)T/8
Hence, total energy is equally shared between the inductor and the capacity at time,
t =T/8,3T/8,5T/8,…
Ans (e).
If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the circuit. The resistance damps out the LC oscillation.
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.
A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply. (a) What is the maximum current in the circuit? (b) What is the time lag between the current maximum and the voltage maximum?
Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F = 10⁻⁴ F Resistance of the resistor, R = 40 Ω Supply voltage, V = 110 V Ans (a). Frequency of oscillations, ν= 60 Hz Angular frequency, ω = 2π ν = 2π x 60 rad/s =120π rad/s For a RC circuit, we have the relation for impedance as: Z = √(R2 +1/ω2Read more
Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F = 10⁻⁴ F
Resistance of the resistor, R = 40 Ω
Supply voltage, V = 110 V
Ans (a).
Frequency of oscillations, ν= 60 Hz
Angular frequency, ω = 2π ν = 2π x 60 rad/s =120π rad/s
For a RC circuit, we have the relation for impedance as: Z = √(R2 +1/ω2c2)
Peak voltage, V0 = V√2 = 110√2. Maximum current is given as:
I0= V0 /Z = V0/√(R2 +1/ω2c2)
=110 √2 / √(R2 +1/ω2c2)
=110 √2 / √(402 +1/(120π)2(10⁻⁴)2)
=3.24A
Ans (b).
In a capacitor circuit, the voltage lags behind the current by a phase angle of φ. This angle is given by the relation:
tan φ =(1/ωC)/R = 1/ωCR
=1/(120 π10⁻⁴x 40)
= 0.6635
Therefore, φ =tan⁻¹ (0.6635) = 33.56º
= 33.56π/180 rad
Therefore ,Time lag = φ/ω = 33.56 π/(180 x 120 π) = 1.55 x 10⁻³ s
=1.55 ms
Hence, the time lag between maximum current and maximum voltage is 1.55 ms.
See lessObtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Inductance of the inductor, L = 0.5 Hz, Resistance of the resistor, R = 100 Ω Potential of the supply voltages, V = 240 V, Frequency of the supply,ν = 10 kHz = 104 Hz Angular frequency, ω = 2πν= 2π x 104 rad/s Ans (a). Peak voltage, V0Read more
Inductance of the inductor, L = 0.5 Hz, Resistance of the resistor, R = 100 Ω
Potential of the supply voltages, V = 240 V,
Frequency of the supply,ν = 10 kHz = 104 Hz
Angular frequency, ω = 2πν= 2π x 104 rad/s
Ans (a).
Peak voltage, V0 = √2 V= 240 √2 V
Maximum Current, I0 = V0 /√ (R² + ω² L²)
=240 √2/√[(100)² + (2π x 104 )² x (0.50)² ]
=1.1 x 10⁻²A
Ans (b).
For phase difference φ,we have the relation :
tanφ = Lω/R =(2π x 104 x 0.5)/100 = 100π
=> φ = 89.82º =89.82π /180 rad
ωt = 89.82π /180
=> t = 89.82π /(180 x 2π x 104) = 25μs
See lessIt can be observed that Io is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.
In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.
An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. (a) What is the total energy stored initially? Is it conserved during LC oscillations? (b) What is the natural frequency of the circuit? (c) At what time is the energy stored (i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e., stored in the inductor)? (d) At what times is the total energy shared equally between the inductor and the capacitor? (e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Inductance of the inductor, L = 20 mH = 20 x 10⁻3 H Capacitance of the capacitor, C = 50 pF = 50 x 10⁻6 F Initial charge on the capacitor, Q = 10 mC = 10 x 10-3 C Ans (a). Total energy stored initially in the circuit is given as: E = 1/2 x Q²/C = (10 x 10-3)²/ (2 x 50 x 10⁻6 ) = 1 J Hence, the totaRead more
Inductance of the inductor, L = 20 mH = 20 x 10⁻3 H
Capacitance of the capacitor, C = 50 pF = 50 x 10⁻6 F
Initial charge on the capacitor, Q = 10 mC = 10 x 10-3 C
Ans (a).
Total energy stored initially in the circuit is given as:
E = 1/2 x Q²/C = (10 x 10-3)²/ (2 x 50 x 10⁻6 ) = 1 J
Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.
Ans (b).
Natural frequency of the circuit is given by the relation,
ν = 1/[2π√(LC)] = 1/(2π√(20 x 10⁻3 x 50 x 10⁻6) = 10³/2π = 159.24 Hz
Natural angular frequency, ωr = 1/√(LC)
= 1/ √(20 x 10⁻3 x 50 x 10⁻6) = 1/ √ 10⁻6= 103 rad/s
Hence, the natural frequency of the circuit is 103 rad/s.
Ans (c).
(i) For time period (T = 1/ν =1/159.24 = 6.28 ms), total charge on the capacitor at time t,
Q’ = Qcos2πt/T
For energy stored is electrical, we can write Q’ = Q.
Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t =0,T/2,T,3T/2
(ii) Magnetic energy is the maximum when electrical energy, Q’ is equal to 0.
Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time, t = T/4 ,3T/4,5T/4—-
Ans (d).
Q1 = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.
When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = 1/2(maximum energy)
=>1/2 x (Q¹)² /C =1/2 x (1/2C x Q²) = Q²/4C
Q¹ =Q/√2
But Q¹ = Q cos(2π/T) .t
=> Q/√2= Q cos(2π/T) .t
=> cos(2π/T) .t= 1/√2 = cos (2n+1)π/4; where n= 0,1,2,…
t= (2n+1)T/8
Hence, total energy is equally shared between the inductor and the capacity at time,
t =T/8,3T/8,5T/8,…
Ans (e).
If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the circuit. The resistance damps out the LC oscillation.
See lessFigure 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 less