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.,
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 less