The rating of a step-down transformer is 40000 V-220 V.
Input voltage,V, = 40000 V
Output voltage,V₂ = 220 V
Total electric power required,
P = 800 kW = 800 x 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
Resistance of the two wire lines carrying power =0.5 Ω/km Total resistance of the wire lines,

R = (15 + 15)0.5 = 15 Ω

P= V1I

Rms current in the wire line is given as:

=> I = P/V1

= 800 x 10³/40000= 20 A

Ans (a).

Line power loss = I²R = (20)² x 15 = 6 kW

Ans (b).

Assuming that the power loss is negligible due to the leakage of current.

Hence, power supplied by the plant = 800 kW + 6kW = 806 kW

Ans (c).

Voltage drop in the power line = IR = 20 x 15 = 300 V

Hence, voltage that is transmitted by the power plant

= 300 + 40000 = 40300 V

The power is being generated in the plant at 440 V.

Hence, the rating of the step-up transformer needed at the plant is 440 V – 40300 V.

Hence, power loss during transmission

= (600/1400 )x 100 = 42.8%

1400

In the previous exercise, the power loss due to the same reason is (6/806 )x 100 = 0.744%

Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.

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

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 = 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

ω_R– 1/ω_RC =0

Where, ω_R  = Resonance angular frequency

Where , ω_R = Resonance angular frequency

Therefore ω_{R = 1/√(LC) = 1/ √ (0.12 x 480 x 10⁻⁹ ) = 4166.67 rad/s}

Resonant frequency , ν_R = ω_R/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:

 (P_av)_Max= 1/2 x  (I₀)²_MaxR

= 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,

=  ω_R ± Δω

= 2π (ν_R ± 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

ν_R +Δν  = 663.48 + 15.26 = 678.74 Hz

and ν_R – 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 = ω_R L/R  = 4166.67 x 0.12 /23 = 21.74

Hence, the Q-factor of the given circuit is 21.74.

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,

V_L= I x ωL

= 8.22 x 100π x 80 x 10⁻³= 206.61 V

Potential difference across the capacitor,

V_c = 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.