#### Topics

1. A 100 Ω resistor is connected to a 220 V, 50 Hz a.c. supply.

(a)  What is the rms value of current in the circuit?

(b) What is the net power consumed over a full cycle?

R = 100 Ω  ,  Erms = 220 volt,  v = 50 Hz

(a) Irms = Erms/R

= 220/100

= 2.2 A

(b) The net power consumed over a full cycle is

P = Erms Irms

= 220 × 2.2

= 484 W

2. (a) The peak voltage of an a.c. supply is 300 V. What is the rms voltage?

(b) The rms value of current in an a.c. circuit is 10 A. What is the peak current?

(a) E0 = 300 V

Erms = E0/√2

= 300/1.414

= 212.16 V

(b) I0 = √2 Irms = 1.414 × 10 = 14.14 A

3. A 44 mH inductor is connected to 220 V, 50 Hz a.c. supply. Determine the rms value of the current in the circuit.

L = 44 × 10-3 H, Erms = 220 V,

v = 50 Hz

4. A 60 μF capacitor is connected to a 110 V, 60 Hz a.c. supply. Determine the rms value of the current in the circuit.

C = 60 μF = 60 × 10-6 F,

Erms = 110 V, v = 60 Hz

5. Obtain the resonant frequency ωr, of a series LCR circuit with L = 2.0 H, C = 32 μF and R =10 Ω. What is the Q - value of this circuit?

Given  L = 2.0 H,  C = 32 μF ,  R = 10 Ω

6. A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

C = 30 μF = 30 × 10-6 F,

L = 27 mH = 27 × 10-3 H

7. A series LCR circuit with R = 20 Ω , L = 1.5 H and C = 35 μF is connected to a variable frequency 200 V a.c. supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

R = 20 Ω , L = 1.5 H,

C = 35 μF = 35 × 10-6 F

Erms = 200 V

When the frequency of the supply equals the natural frequency of circuit, resonance occurs.

At resonance, Z = R and

Therefore , average power transferred/cycle

P = Erms Irms cos Φ = 200 ×10 × cos 0o = 2000 W

8. 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 μH, 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]

Given: L = 200 μH = 200 × 10-6 H

For frequency 800 KHz

For frequency 1200 KHz

9. A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz a.c. supply.

(a) What is the maximum current in the coil?

(b) What is the time lag between the voltage maximum and the current maximum?

(a) f = 50 Hz, L = 0.50 H, R =100 Ω

Ev = 240 V

(b) For an RL circuit

If E = E0 sin ωt

I = I0 sin (ωt - Φ )

E will be maximum when

ωt = π/2

and I will be maximum when

(ωt - Φ ) = π/2

The phase difference between the voltage and current is given by

10. 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?

Given  C = 100 µF = 100 × 10-6 F = 10-4 F

R = 40 Ω , Erms = 110 V, f = 60 Hz

11. Obtain the resonant frequency and Q - factor of series LCR circuit with L = 3.0 H, C = 27 µF, and R = 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ' full with at half maximum ' by a factor of 2. Suggest a suitable way.

Given  L = 3.0 H,  C = 27 µF = 27 × 10-6 F,

R = 7.4 Ω

Resonant frequency

To double the value of Q, R should be halved , i.e., it should be 3.7 Ω. It is to be noted that ωrL is not be changed as reducing the full width at half maximum implies that resonant frequency (ωr) is not be disturbed.

12. At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3s-1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms-1).

Height of water h = 300 m

Rate of volume flow of water

= dV/dt = 100 m3s-1

Turbine generator efficiency

η = 60% = 0.6

Potential Energy of water of mass m at height h, i.e.,

W = mgh

Hydroelectric power Ph = d/dt (mgh)

Ph = d/dt (Vρgh) = ρgh dV/dt  ( m = Vρ)

Ph = (1000 kgm-3) (9.8 ms-2) (300 m) (100 m3s-1)

= 29.4 × 107 W

Electric power = η × Ph = 0.6 × 29.4 × 107

= 17.64 × 107 W

13. 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 4000 - 220 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) Characterize the step up transformer at the plant.

(a) The resistance of the line of 15 km is

R = (15 + 15) × 0.5 = 15 Ω

The line power loss = I2rms R

= (200)2 × 15 = 6 × 105 W

(b) The power supplied by the plant

= 8 × 105 + 6 × 105 = 14 × 105 W

(c) The voltage drop across the line

= Irms R = 200 × 15 = 3000 V

Therefore , the step up transformer at the plant

= [ 440 V, (3000 + 4000) V ]

= [ 400 V, 7000 V ]

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