Total electric power required, P = 800 kW = 800 x 103 W Supply voltage, V = 220 V Voltage at which electric plant is generating power, V' = 440 V Distance between the town and power generating station, d = 15 km Resistance of the two wire lines carrying power =0.5 Ω/km Total resistance of the wires,Read more
Total electric power required, P = 800 kW = 800 x 103 W
Supply voltage, V = 220 V
Voltage at which electric plant is generating power, V’ = 440 V
Distance between the town and power generating station, d = 15 km
Resistance of the two wire lines carrying power =0.5 Ω/km
Total resistance of the wires, R = (15 + 15)0.5 =15Ω
A step-down transformer of rating 4000 – 220 V is used in the sub-station.
Input voltage, V1 = 4000 V
Output voltage, V2 = 220 V
rms current in the wire lines is given as:
I = P V1 = 800 x 10³/4000 = 200 A
Ans (a).
Line power loss = I2R = (200)2 x 15 =600 x103W = 600 kW
Ans (b).
Assuming that the power loss is negligible due to the leakage of the current:
Total power supplied by the plant = 800 kW + 600 kW = 1400 kW
Ans (c).
Voltage drop in the power line = IR = 200 x 15 = 3000 V
Hence, total voltage transmitted from the plant
= 3000 + 4000 =7000 V
Also, the power generated is 440 V.
Hence, the rating of the step-up transformer situated at the power plant is 440 V — 7000 V.
Height of water pressure head, h = 300 m Volume of water flow per second, V = 100 m3/s Efficiency of turbine generator, n = 60% = 0.6 Acceleration due to gravity, g = 9.8 m/s2 Density of water, ρ = 103 kg/m3 Electric power available from the plant = η x hρgV = 0.6 x 300 x 103 x 9.8 x 100 = 176.4 x 1Read more
Height of water pressure head, h = 300 m Volume of water flow per second, V = 100 m3/s Efficiency of turbine generator, n = 60% = 0.6 Acceleration due to gravity, g = 9.8 m/s2 Density of water, ρ = 103 kg/m3 Electric power available from the plant = η x hρgV = 0.6 x 300 x 103 x 9.8 x 100 = 176.4 x 106 W = 176.4 MW
Input voltage, V1 = 2300 V Number of turns in primary coil, n1 = 4000 Output voltage, V2 = 230 V Number of turns in secondary coil = n2 Voltage is related to the number of turns as: V1/V2=n1/n2 => 2300/230 = 4000/n2 => n2= 4000 x 230 /2300 = 400 Hence, there are 400 turns in the second windingRead more
Ans (a). Yes; the statement is not true for rms voltage It is true that in any ac circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltage because voltages across different elements mayRead more
Ans (a).
Yes; the statement is not true for rms voltage
It is true that in any ac circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltage because voltages across different elements may not be in phase.
Ans (b).
High induced voltage is used to charge the capacitor.
A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.
Ans (c).
The dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is negligible while the impedance of a capacitor (C) is very high (almost infinite). Hence, a dc signal appears across C. For an ac signal of high frequency, the impedance of L is high and that of C is very low. Hence, an ac signal of high frequency appears across L.
Ans (d).
If an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), then the lamp will glow dimly. This is because the choke coil and the iron core increase the impedance of the circuit.
Ans (e).
A choke coil is needed in the use of fluorescent tubes with ac mains because it reduces the voltage across the tube without wasting much power. An ordinary resistor cannot be used instead of a choke coil for this purpose because it wastes power in the form of heat.
Inductance, L = 3.0 H, Capacitance, C = 27 μF = 27 x 10-6 F, Resistance, R = 7.4 Ω At resonance, angular frequency of the source for the given LCR series circuit is given as ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s Q-factor of the series: Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446 To improve theRead more
Inductance, L = 3.0 H, Capacitance, C = 27 μF = 27 x 10-6 F, Resistance, R = 7.4 Ω
At resonance, angular frequency of the source for the given LCR series circuit is given as
ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s
Q-factor of the series:
Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446
To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changing ωr, we need to reduce R to half i.e.,
Inductance, L = 0.12 H Capacitance, C = 480 nF = 480 x 10⁻9 F Resistance, R = 23 Ω Supply voltage, V = 230 V Peak voltage is given as: Vo =√2 x 230 = 325.22 V Ans (a). Current flowing in the circuit is given by the relation, I0= V0/√[(R2 + (ωL- 1/ωC)2] Where, Io = maximum at resonance At resonance,Read more
Inductance, L = 0.12 H
Capacitance, C = 480 nF = 480 x 10⁻9 F
Resistance, R = 23 Ω
Supply voltage, V = 230 V
Peak voltage is given as: Vo =√2 x 230 = 325.22 V
Ans (a).
Current flowing in the circuit is given by the relation,
I0= V0/√[(R2 + (ωL- 1/ωC)2]
Where, Io = maximum at resonance
At resonance, we have
ωR– 1/ωRC =0
Where, ωR = Resonance angular frequency
Where , ωR = Resonance angular frequency
Therefore ωR = 1/√(LC) = 1/ √ (0.12 x 480 x 10⁻9 ) = 4166.67 rad/s
Resonant frequency , νR = ωR/2π = 4166.67 /(2 x 3.14) = 663.48 Hz
And ,maximum current, (I0)Max = V0/R = 325.22/23 = 14.14 A
Ans (b).
Maximum average power absorbed by the circuit is given as:
(Pav)Max= 1/2 x (I0)²MaxR
= 1/2 x (14.14)² x 23 = 2299.3 W
Hence, resonant frequency is 663.48 Hz.
Ans (c).
The power transferred to the circuit is half the power at resonant frequency.
Frequencies at which power transferred is half,
= ωR ± Δω
= 2π (νR ± Δν)
Where,
Δω=R/2L= 23/(2 x 0.12)= 95.83 rad/s
Hence, change in frequency,
Δν = 1/2π x Δω =95.83 /2π = 15.26 Hz
Therefore
νR +Δν = 663.48 + 15.26 = 678.74 Hz
and νR – Δν = 663.48 – 15.26 =648.22 Hz
Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half. At these frequencies, current amplitude can be given as:
I’ = 1/√ 2 x (I0)Max
= 14.14/√ 2 = 10A
Ans (d).
Q-factor of the given circuit can be obtained using the relation,
Q = ωR L/R = 4166.67 x 0.12 /23 = 21.74
Hence, the Q-factor of the given circuit is 21.74.
Average power transferred to the resistor = 788.44 W Average power transferred to the capacitor = 0 W Total power absorbed by the circuit = 788.44 W Inductance of inductor, L = 80 mH = 80 x 10-3 H Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Resistance of resistor, R = 15 Ω Potential of voltageRead more
Average power transferred to the resistor = 788.44 W Average power transferred to the capacitor = 0 W Total power absorbed by the circuit = 788.44 W Inductance of inductor, L = 80 mH = 80 x 10-3 H Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Resistance of resistor, R = 15 Ω Potential of voltage supply, V = 230 V Frequency of signal, ν = 50 Hz Angular frequency of signal, ω = 2πν= 2πx (50) = 100π rad/s The elements are connected in series to each other. Hence, impedance of the circuit is given as: Z = √[(R)² +(ωL -1/ωC)²]
=√[15² +((100π x 80 x 10-3 )-1/(100πx 60 x 10-6 ))²]
= √[(15)² +(25.12 -53.08)²] = 31.728 Ω
Current flowing in the circuit, I = V/Z = 230/31.728 = 7.25 A
Average power transferred to resistance is given as:
PR = I2R = (7.25)2 x 15 = 788.44 W
Average power transferred to capacitor, PC = Average power transferred to inductor, PL= 0
Total power absorbed by the circuit: = PR + PC + PL = 788.44 + 0 + 0 = 788.44 W
Hence, the total power absorbed by the circuit is 788.44 W.
Inductance, L = 80 mH = 80 x 10⁻3 H, Capacitance, C = 60 pF = 60 x 10⁻6 F Supply voltage, V = 230 V, Frequency, ν = 50 Hz, Angular frequency, ω = 2πν = 100 π rad/s Peak voltage, Vo= √2 = 230√2 V Ans (a). Maximum current is given as : Io = Vo /(1/ωL -ωC) = 230 √3 /[(100π x 80 x 10⁻3 )- 1/ (100π x 60Read more
Inductance, L = 80 mH = 80 x 10⁻3 H, Capacitance, C = 60 pF = 60 x 10⁻6 F
Supply voltage, V = 230 V,
Frequency, ν = 50 Hz,
Angular frequency, ω = 2πν = 100 π rad/s
Peak voltage, Vo= √2 = 230√2 V
Ans (a).
Maximum current is given as :
Io = Vo /(1/ωL -ωC)
= 230 √3 /[(100π x 80 x 10⁻3 )- 1/ (100π x 60 x 10⁻6 )]
= 230√2 / (8π -1000/6π) = -11.63 A
The negative sign appears because ωL <1/ωC
Amplitude of maximum current, |I0| = 11.63 A
Hence, rms value of current =I = I0/√2 =11.63 /√2 = -8.22 A
Ans (b).
Potential difference across the inductor,
VL= I x ωL
= 8.22 x 100π x 80 x 10⁻3= 206.61 V
Potential difference across the capacitor,
Vc = I x 1/ωC
= 8.22 x 1/(100π x 60 x 10⁻6 ) = 436.3 V
Ans (c).
Average power consumed by the inductor is zero as actual voltage leads the current by π/2
Ans (d).
Average power consumed by the capacitor is zero as voltage lags current by π/2
Ans (e).
The total power absorbed (averaged over one cycle) is zero.
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where, L = 5.0 H, C = 80 μF = 80 x 10-6F, R = 40Ω Potential of the voltage source, V = 230 V IRead more
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,
L = 5.0 H, C = 80 μF = 80 x 10–6F, R = 40Ω
Potential of the voltage source, V = 230 V Impedance (Z) of the given parallel LCR circuit is given as:
1/Z = √(1/R² +(1/ωL -ωC)²)
Where, ω = Angular frequency Z
At resonance, (1/ωL – ωC) = 0
Therefore, ω = 1/√(LC) = 1 / √(5 x 80 x 10–6) = 50 rad/s
Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum, rms current flowing through inductor L is given as:
IL= V/ωL = 230 /(50 x 5) = 0.92 A
rms current flowing through capacitor C is given as :
IC = V/(1/ωC) =ωCV = 50 x 80 x 10–6 x 230 =0.92A
rms current flowing through resistor R is given as :
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.
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω. per km. The town gets power from the line through a 4000220 V step-down transformer at a sub-station in the town. (a) Estimate the line power loss in the form of heat. (b) How much power must the plant supply, assuming there is negligible power loss due to leakage? (c) Characterise the step up transformer at the plant.
Total electric power required, P = 800 kW = 800 x 103 W Supply voltage, V = 220 V Voltage at which electric plant is generating power, V' = 440 V Distance between the town and power generating station, d = 15 km Resistance of the two wire lines carrying power =0.5 Ω/km Total resistance of the wires,Read more
Total electric power required, P = 800 kW = 800 x 103 W
Supply voltage, V = 220 V
Voltage at which electric plant is generating power, V’ = 440 V
Distance between the town and power generating station, d = 15 km
Resistance of the two wire lines carrying power =0.5 Ω/km
Total resistance of the wires, R = (15 + 15)0.5 =15Ω
A step-down transformer of rating 4000 – 220 V is used in the sub-station.
Input voltage, V1 = 4000 V
Output voltage, V2 = 220 V
rms current in the wire lines is given as:
I = P V1 = 800 x 10³/4000 = 200 A
Ans (a).
Line power loss = I2R = (200)2 x 15 =600 x103W = 600 kW
Ans (b).
Assuming that the power loss is negligible due to the leakage of the current:
Total power supplied by the plant = 800 kW + 600 kW = 1400 kW
Ans (c).
Voltage drop in the power line = IR = 200 x 15 = 3000 V
Hence, total voltage transmitted from the plant
= 3000 + 4000 =7000 V
Also, the power generated is 440 V.
Hence, the rating of the step-up transformer situated at the power plant is 440 V — 7000 V.
See lessAt a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m³ s⁻¹. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms⁻²).
Height of water pressure head, h = 300 m Volume of water flow per second, V = 100 m3/s Efficiency of turbine generator, n = 60% = 0.6 Acceleration due to gravity, g = 9.8 m/s2 Density of water, ρ = 103 kg/m3 Electric power available from the plant = η x hρgV = 0.6 x 300 x 103 x 9.8 x 100 = 176.4 x 1Read more
Height of water pressure head, h = 300 m
See lessVolume of water flow per second, V = 100 m3/s
Efficiency of turbine generator, n = 60% = 0.6
Acceleration due to gravity, g = 9.8 m/s2
Density of water, ρ = 103 kg/m3
Electric power available from the plant = η x hρgV
= 0.6 x 300 x 103 x 9.8 x 100 = 176.4 x 106 W = 176.4 MW
A power transmission line feeds input power at 2300 V to a stepdown transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Input voltage, V1 = 2300 V Number of turns in primary coil, n1 = 4000 Output voltage, V2 = 230 V Number of turns in secondary coil = n2 Voltage is related to the number of turns as: V1/V2=n1/n2 => 2300/230 = 4000/n2 => n2= 4000 x 230 /2300 = 400 Hence, there are 400 turns in the second windingRead more
Input voltage, V1 = 2300 V
Number of turns in primary coil, n1 = 4000
Output voltage, V2 = 230 V
Number of turns in secondary coil = n2
Voltage is related to the number of turns as:
V1/V2=n1/n2
=> 2300/230 = 4000/n2
=> n2= 4000 x 230 /2300 = 400
Hence, there are 400 turns in the second winding.
See lessAnswer the following questions: (a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage? (b) A capacitor is used in the primary circuit of an induction coil. (c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L. (d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line. (e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
Ans (a). Yes; the statement is not true for rms voltage It is true that in any ac circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltage because voltages across different elements mayRead more
Ans (a).
Yes; the statement is not true for rms voltage
It is true that in any ac circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltage because voltages across different elements may not be in phase.
Ans (b).
High induced voltage is used to charge the capacitor.
A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.
Ans (c).
The dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is negligible while the impedance of a capacitor (C) is very high (almost infinite). Hence, a dc signal appears across C. For an ac signal of high frequency, the impedance of L is high and that of C is very low. Hence, an ac signal of high frequency appears across L.
Ans (d).
If an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), then the lamp will glow dimly. This is because the choke coil and the iron core increase the impedance of the circuit.
Ans (e).
A choke coil is needed in the use of fluorescent tubes with ac mains because it reduces the voltage across the tube without wasting much power. An ordinary resistor cannot be used instead of a choke coil for this purpose because it wastes power in the form of heat.
See lessObtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ωμ. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘fiill width at half maximum’ by a factor of 2. Suggest a suitable way.
Inductance, L = 3.0 H, Capacitance, C = 27 μF = 27 x 10-6 F, Resistance, R = 7.4 Ω At resonance, angular frequency of the source for the given LCR series circuit is given as ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s Q-factor of the series: Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446 To improve theRead more
Inductance, L = 3.0 H,
Capacitance, C = 27 μF = 27 x 10-6 F,
Resistance, R = 7.4 Ω
At resonance, angular frequency of the source for the given LCR series circuit is given as
ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s
Q-factor of the series:
Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446
To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changing ωr, we need to reduce R to half i.e.,
Resistance = R/2 = 7.4/2 = 3.7 Ω
See lessA series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum? Obtain this maximum value. (b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain die value of this maximum power. (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? (d) What is the Q-factor of the given circuit?
Inductance, L = 0.12 H Capacitance, C = 480 nF = 480 x 10⁻9 F Resistance, R = 23 Ω Supply voltage, V = 230 V Peak voltage is given as: Vo =√2 x 230 = 325.22 V Ans (a). Current flowing in the circuit is given by the relation, I0= V0/√[(R2 + (ωL- 1/ωC)2] Where, Io = maximum at resonance At resonance,Read more
Inductance, L = 0.12 H
Capacitance, C = 480 nF = 480 x 10⁻9 F
Resistance, R = 23 Ω
Supply voltage, V = 230 V
Peak voltage is given as: Vo =√2 x 230 = 325.22 V
Ans (a).
Current flowing in the circuit is given by the relation,
I0= V0/√[(R2 + (ωL- 1/ωC)2]
Where, Io = maximum at resonance
At resonance, we have
ωR– 1/ωRC =0
Where, ωR = Resonance angular frequency
Where , ωR = Resonance angular frequency
Therefore ωR = 1/√(LC) = 1/ √ (0.12 x 480 x 10⁻9 ) = 4166.67 rad/s
Resonant frequency , νR = ωR/2π = 4166.67 /(2 x 3.14) = 663.48 Hz
And ,maximum current, (I0)Max = V0/R = 325.22/23 = 14.14 A
Ans (b).
Maximum average power absorbed by the circuit is given as:
(Pav)Max= 1/2 x (I0)²MaxR
= 1/2 x (14.14)² x 23 = 2299.3 W
Hence, resonant frequency is 663.48 Hz.
Ans (c).
The power transferred to the circuit is half the power at resonant frequency.
Frequencies at which power transferred is half,
= ωR ± Δω
= 2π (νR ± Δν)
Where,
Δω=R/2L= 23/(2 x 0.12)= 95.83 rad/s
Hence, change in frequency,
Δν = 1/2π x Δω =95.83 /2π = 15.26 Hz
Therefore
νR +Δν = 663.48 + 15.26 = 678.74 Hz
and νR – Δν = 663.48 – 15.26 =648.22 Hz
Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half. At these frequencies, current amplitude can be given as:
I’ = 1/√ 2 x (I0)Max
= 14.14/√ 2 = 10A
Ans (d).
Q-factor of the given circuit can be obtained using the relation,
Q = ωR L/R = 4166.67 x 0.12 /23 = 21.74
Hence, the Q-factor of the given circuit is 21.74.
See lessSuppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Average power transferred to the resistor = 788.44 W Average power transferred to the capacitor = 0 W Total power absorbed by the circuit = 788.44 W Inductance of inductor, L = 80 mH = 80 x 10-3 H Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Resistance of resistor, R = 15 Ω Potential of voltageRead more
Average power transferred to the resistor = 788.44 W
Average power transferred to the capacitor = 0 W
Total power absorbed by the circuit = 788.44 W
Inductance of inductor, L = 80 mH = 80 x 10-3 H
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F
Resistance of resistor, R = 15 Ω
Potential of voltage supply, V = 230 V
Frequency of signal, ν = 50 Hz
Angular frequency of signal, ω = 2 πν= 2π x (50) = 100π rad/s
The elements are connected in series to each other. Hence, impedance of the circuit is given as:
Z = √[(R)² +(ωL -1/ωC)²]
=√[15² +((100π x 80 x 10-3 )-1/(100πx 60 x 10-6 ))²]
= √[(15)² +(25.12 -53.08)²] = 31.728 Ω
Current flowing in the circuit, I = V/Z = 230/31.728 = 7.25 A
Average power transferred to resistance is given as:
PR = I2R = (7.25)2 x 15 = 788.44 W
Average power transferred to capacitor, PC = Average power transferred to inductor, PL= 0
Total power absorbed by the circuit: = PR + PC + PL = 788.44 + 0 + 0 = 788.44 W
Hence, the total power absorbed by the circuit is 788.44 W.
See lessA circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. (a) Obtain the current amplitude and mis values. (b) Obtain the rms values of potential drops across each element. (c) What is the average power transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]
Inductance, L = 80 mH = 80 x 10⁻3 H, Capacitance, C = 60 pF = 60 x 10⁻6 F Supply voltage, V = 230 V, Frequency, ν = 50 Hz, Angular frequency, ω = 2πν = 100 π rad/s Peak voltage, Vo= √2 = 230√2 V Ans (a). Maximum current is given as : Io = Vo /(1/ωL -ωC) = 230 √3 /[(100π x 80 x 10⁻3 )- 1/ (100π x 60Read more
Inductance, L = 80 mH = 80 x 10⁻3 H,
Capacitance, C = 60 pF = 60 x 10⁻6 F
Supply voltage, V = 230 V,
Frequency, ν = 50 Hz,
Angular frequency, ω = 2πν = 100 π rad/s
Peak voltage, Vo= √2 = 230√2 V
Ans (a).
Maximum current is given as :
Io = Vo /(1/ωL -ωC)
= 230 √3 /[(100π x 80 x 10⁻3 )- 1/ (100π x 60 x 10⁻6 )]
= 230√2 / (8π -1000/6π) = -11.63 A
The negative sign appears because ωL <1/ωC
Amplitude of maximum current, |I0| = 11.63 A
Hence, rms value of current =I = I0/√2 =11.63 /√2 = -8.22 A
Ans (b).
Potential difference across the inductor,
VL= I x ωL
= 8.22 x 100π x 80 x 10⁻3= 206.61 V
Potential difference across the capacitor,
Vc = I x 1/ωC
= 8.22 x 1/(100π x 60 x 10⁻6 ) = 436.3 V
Ans (c).
Average power consumed by the inductor is zero as actual voltage leads the current by π/2
Ans (d).
Average power consumed by the capacitor is zero as voltage lags current by π/2
Ans (e).
The total power absorbed (averaged over one cycle) is zero.
See lessKeeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where, L = 5.0 H, C = 80 μF = 80 x 10-6F, R = 40Ω Potential of the voltage source, V = 230 V IRead more
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,
L = 5.0 H, C = 80 μF = 80 x 10–6F, R = 40Ω
Potential of the voltage source, V = 230 V Impedance (Z) of the given parallel LCR circuit is given as:
1/Z = √(1/R² +(1/ωL -ωC)²)
Where, ω = Angular frequency Z
At resonance, (1/ωL – ωC) = 0
Therefore, ω = 1/√(LC) = 1 / √(5 x 80 x 10–6) = 50 rad/s
Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum, rms current flowing through inductor L is given as:
IL= V/ωL = 230 /(50 x 5) = 0.92 A
rms current flowing through capacitor C is given as :
IC = V/(1/ωC) =ωCV = 50 x 80 x 10–6 x 230 =0.92A
rms current flowing through resistor R is given as :
IR= V/R = 230/40 = 5.75A
See lessObtain 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.
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