Impedance of the circuit is given by the relation,
Current in the circuit can be calculated as:
Hence, the average power transferred to the circuit in one complete cycle= VI
= 200 × 10 = 2000 W.
A 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 200 ÃŽÂ¼H, 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 radiowave.]
The range of frequency (ÃŽÂ½) of a radio is 800 kHz to 1200 kHz.
Hence, the time lag between maximum voltage and maximum current is.
Now, phase angle Φis given by the relation,
Hence, the time lag between maximum voltage and maximum current is 3.2 ms.
Obtain 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?
For a RC circuit, we have the relation for impedance as:
Peak voltage, V0 =
Maximum current is given as:
(b) In a capacitor circuit, the voltage lags behind the current by a phase angle ofΦ. This angle is given by the relation:
Hence, the time lag between maximum current and maximum voltage is 1.55 ms.
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 ÃŽÂ¼F = 100 × 10 - 6 F
For an RC circuit, the voltage lags behind the current by a phase angle of Φ given as:
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, Ãâ€° = 0. Hence, capacitor C amounts to an open circuit.
Keeping 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 parallelLCR 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,
At resonance, angular frequency of the source for the given LCR series circuit is given as:
Q-factor of the series:
To improve the sharpness of the resonance by reducing its Ã¢â‚¬Ëœfull width at half maximum' by a factor of 2 without changing, we need to reduce R to half i.e.,
Answer 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?
(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.
(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.
(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 Cis very low. Hence, an ac signal of high frequency appears across L.
(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.
(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.
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
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:
Hence, there are 400 turns in the second winding.
At 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 m s-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, ÃÂ = 103 kg/m3
Electric power available from the plant = ÃŽÂ· x hÃÂgV
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:
(a) Line power loss = I2R
= (200)2 × 15
= 600 × 103 W
= 600 kW
(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
(c) Voltage drop in the power line = IR = 200 × 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.
Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?
The rating of a step-down transformer is 40000 V - 220 V.
Input voltage, V1 = 40000 V
Output voltage, V2 = 220 V
Total electric power required, P = 800 kW = 800 × 103 W
Source potential, V = 220 V
Voltage at which the electric plant generates power, V' = 440 V
Distance between the town and power generating station, d = 15 km