Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F =10⁻⁴F Resistance of the resistor, R = 40Ω Supply voltage, V = 110 V Frequency of the supply, ν = 12 kHz = 12 x 103 Hz Angular Frequency, ω = 2π ν = 2 x π x 12 x 103 Hz = 24π x 103 rad/s Peak voltage, V0 = V√2 = 110√2 V Maximum Current ,I0= VRead more
Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F =10⁻⁴F Resistance of the resistor, R = 40Ω Supply voltage, V = 110 V Frequency of the supply, ν = 12 kHz = 12 x 103 Hz Angular Frequency, ω = 2π ν = 2 x π x 12 x 103 Hz = 24π x 103 rad/s Peak voltage, V0 = V√2 = 110√2 V
Maximum Current ,I0= V0/√(R2 +1/ω2c2)
=110 √2 /√[(40)2 +1/(24π x 103)2(10⁻⁴)2]
=110 √2/√[1600 + (10/24π)2] = 3.9 A
For an RC circuit, the voltage lags behind the current by a phase angle of φ given as:
tan φ = (1/ωC)/R = 1/ωCR = 1/(24π x 103 x 10⁻⁴ x 40)
tan φ = 1/96π
Therefore , φ = 0.2º or 0.2 π/180 rad
Therefore ,Time lag = φ/ω = 0.2 π/(180 x 24π x 103)
=4.69 x 10⁻³s
= 0.04μs
Hence, φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor. In a dc circuit, after the steady state is achieved, ω = 0.
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 = 0.50 H Resistance of the resistor, R = 100Ω. Potential of the supply voltage, V = 240 V Frequency of the supply, v = 50 Hz Ans (a). Peak voltage is given as: V0 = √2 V = √2 x 240= 339.41 V Angular frequency of the supply, ω = 2πν = 2π x 50 = 100π rad/s Maximum currentRead more
Inductance of the inductor, L = 0.50 H
Resistance of the resistor, R = 100Ω.
Potential of the supply voltage, V = 240 V
Frequency of the supply, v = 50 Hz
Ans (a).
Peak voltage is given as: V0 = √2 V
= √2 x 240= 339.41 V
Angular frequency of the supply, ω = 2πν = 2π x 50 = 100π rad/s
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.
Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F =10⁻⁴F Resistance of the resistor, R = 40Ω Supply voltage, V = 110 V Frequency of the supply, ν = 12 kHz = 12 x 103 Hz Angular Frequency, ω = 2π ν = 2 x π x 12 x 103 Hz = 24π x 103 rad/s Peak voltage, V0 = V√2 = 110√2 V Maximum Current ,I0= VRead more
Capacitance of the capacitor, C = 100 μF = 100 x 10-6 F =10⁻⁴F
Resistance of the resistor, R = 40Ω
Supply voltage, V = 110 V
Frequency of the supply, ν = 12 kHz = 12 x 103 Hz
Angular Frequency, ω = 2π ν = 2 x π x 12 x 103 Hz
= 24π x 103 rad/s
Peak voltage, V0 = V√2 = 110√2 V
Maximum Current ,I0= V0/√(R2 +1/ω2c2)
=110 √2 /√[(40)2 +1/(24π x 103)2(10⁻⁴)2]
=110 √2/√[1600 + (10/24π)2] = 3.9 A
For an RC circuit, the voltage lags behind the current by a phase angle of φ given as:
tan φ = (1/ωC)/R = 1/ωCR = 1/(24π x 103 x 10⁻⁴ x 40)
tan φ = 1/96π
Therefore , φ = 0.2º or 0.2 π/180 rad
Therefore ,Time lag = φ/ω = 0.2 π/(180 x 24π x 103)
=4.69 x 10⁻³s
= 0.04μs
Hence, φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor. In a dc circuit, after the steady state is achieved, ω = 0.
Hence, capacitor C amounts to an open circuit.
See lessA 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.
A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum?
Inductance of the inductor, L = 0.50 H Resistance of the resistor, R = 100Ω. Potential of the supply voltage, V = 240 V Frequency of the supply, v = 50 Hz Ans (a). Peak voltage is given as: V0 = √2 V = √2 x 240= 339.41 V Angular frequency of the supply, ω = 2πν = 2π x 50 = 100π rad/s Maximum currentRead more
Inductance of the inductor, L = 0.50 H
Resistance of the resistor, R = 100Ω.
Potential of the supply voltage, V = 240 V
Frequency of the supply, v = 50 Hz
Ans (a).
Peak voltage is given as: V0 = √2 V
= √2 x 240= 339.41 V
Angular frequency of the supply, ω = 2πν = 2π x 50 = 100π rad/s
Maximum current in the circuit is given as:
I0 = V0 /√ (R² + ω² L²) =339.41/√[ (100)² + (100π)² (0.50)²
=1.82 A
Ans (b).
Equation for voltage is given as : V= V0 cosωt
Equation for current is given as ;I = I0 cos(ωt-φ )
Where,φ = Phase difference between voltage and current .At time t=0
V= V0(voltage is maximum)
Hence, the time lag between maximum voltage and maximum current is φ/ω
Now, phase angle φ is given by the relation,
tan φ = ωL/R
=(2π x 50 x 0.5)/100=1.57
=>φ= 57.5º = 57.5 π/180 rad
=>ωt =57.5 π/180
=>t = 57.5 /(180 x 2π x50)= 3.19 x 10⁻³ s = 3.2ms
Hence ,the time lag between maximum voltage and maximum current is 3.2 ms
See lessAn 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 less