(a) (i) What is meant by armature reaction in DC machines ? Show with the help of developed view of armature conductors and poles that the effect of armature m.m.f. on the main field is entirely cross-magnetizing. (10 marks)

(ii) A 10 kW, 220 V DC s

Question

(a) (i) What is meant by armature reaction in DC machines ? Show with the help of developed view of armature conductors and poles that the effect of armature m.m.f. on the main field is entirely cross-magnetizing. (10 marks)

(ii) A 10 kW, 220 V DC shunt motor draws a line current of 5 A while running at no-load speed of 1200 rpm. It has an armature resistance of 0·2 Ω and field resistance of 200 Ω. Determine the efficiency of the motor when it delivers rated load. (10 marks)

(b) A converter circuit as shown in the figure is being used to charge a battery of voltage E = 24 V. The average charging current I_dc = 6 A, and supply voltage V_s = 60 V, 50 Hz. Determine (i) the value of limiting resistor 'R', and (ii) input power factor. (20 marks)

(c) A DSB-SC amplitude-modulated signal with power spectral density as shown in figure (a) is corrupted with additive noise that has a power spectral density (N_0/2) within the passband region of the signal. The received signal-plus-noise is demodulated and low pass filtered as shown in figure (b). Determine the SNR at the output of the LPF. [BW : bandwidth] [Given : carrier signal = cos (2πf_c t)] (20 marks)

UPSC Answer Check · Accepted Answer

(a)(i) Armature reaction is the effect of the magnetic field set up by the armature current on the main field flux of a DC machine. In a developed view, place the N and S main poles on the stator and the armature conductors below them. With brushes on the geometrical neutral axis (GNA), the current in all armature conductors under one pole flows in one direction, and under the adjacent pole it reverses. Applying the right-hand rule, the armature m.m.f. wave is directed along the brush axis, i.e. the interpolar or q-axis. The main field m.m.f. is along the pole axis, i.e. the d-axis. Since these two axes are perpendicular, the armature m.m.f. has no component along the main field axis. Its effect is to strengthen one pole tip and weaken the other, distorting the main field and shifting the neutral axis. Hence, for brushes on the GNA, the armature reaction is entirely cross-magnetizing. Only if the brushes are shifted from the GNA does a demagnetizing or magnetizing component appear.

(a)(ii)
Field current, I_f = V/R_f = 220/200 = 1.1 A.
No-load armature current, I_a0 = I_L0 − I_f = 5 − 1.1 = 3.9 A.
Field Cu loss = V I_f = 220 × 1.1 = 242 W.
No-load armature Cu loss = I_a0² R_a = 3.9² × 0.2 = 3.042 W.
No-load input = 220 × 5 = 1100 W.
Constant losses, P_c = 1100 − 242 − 3.042 = 854.958 W.

At rated load, P_out = 10 kW = 10000 W. Let I_a be the rated armature current. Power balance gives:
10000 = 220 I_a − 0.2 I_a² − 854.958
So, 0.2 I_a² − 220 I_a + 10854.958 = 0.
Solving, I_a = [220 ± √(220² − 4 × 0.2 × 10854.958)]/(2 × 0.2)
= [220 ± √39716.034]/0.4.
Taking the lower root, I_a = 51.778 A.

Rated armature Cu loss = 0.2 × 51.778² = 536.19 W.
Total losses = 854.958 + 242 + 536.19 = 1633.15 W.
Input at rated load = 10000 + 1633.15 = 11633.15 W.
Efficiency, η = (10000/11633.15) × 100 = 85.96%.

η ≈ 85.96%.

(b)(i) The circuit is a single-phase half-wave diode battery charger. Taking the AC supply voltage as rms,
V_m = √2 × 60 = 84.853 V.
Conduction starts when V_m sinθ = E:
α = sin⁻¹(E/V_m) = sin⁻¹(24/84.853) = sin⁻¹(√2/5) = 0.286756 rad.
Conduction interval is α to π − α.
Average current:
I_dc = (1/2πR) ∫_α^(π−α) (V_m sinθ − E) dθ
= [2V_m cosα − E(π − 2α)]/(2πR).

Here, V_m cosα = 60√2 × √23/5 = 12√46 = 81.388 V.
π − 2α = 2.56808 rad.
Numerator = 2 × 81.388 − 24 × 2.56808 = 101.142 V·rad.
Given I_dc = 6 A:
R = 101.142/(2π × 6) = 2.683 Ω.

R ≈ 2.683 Ω.

(b)(ii) The rms current is found from
I_rms² = (1/2πR²) ∫_α^(π−α) (V_m sinθ − E)² dθ.
Let J = ∫_α^(π−α) (V_m sinθ − E)² dθ
= (V_m²/2 + E²)(π − 2α) + (V_m²/2) sin2α − 4 V_m E cosα.

With V_m² = 7200, E² = 576, sin2α = 2√46/25 = 0.542586:
J = (3600 + 576)(2.56808) + 3600 × 0.542586 − 4 × 84.853 × 24 × 0.959166
= 4864.37 V²·rad.

Thus, I_rms² = 4864.37/(2π × 2.6829²) = 107.56 A².
I_rms = 10.371 A.

Average input power,
P_in = E I_dc + R I_rms²
= 24 × 6 + 2.6829 × 107.56
= 432.57 W.

Input power factor = P_in/(V_s I_rms)
= 432.57/(60 × 10.371) = 0.695.

Input power factor ≈ 0.695.

(c) For the DSB-SC signal,
Su(f) = (1/4)[S_m(f − f_c) + S_m(f + f_c)].
The positive-frequency lobe extends from f_c − W to f_c + W and consists of two triangles. Its area is
2 × (1/2 × W × P₀) = P₀W.
Hence the message power is P_m = 4P₀W.

After coherent demodulation by cos(2πf_c t) and ideal LPF with BW = W, the output signal is 0.5 m(t).
Output signal power,
P_s,out = (1/4)P_m = P₀W.

The input noise PSD is N₀/2 in both passbands. Multiplication by cos(2πf_c t) gives, for |f| ≤ W,
S_n,out(f) = (1/4)[N₀/2 + N₀/2] = N₀/4.
Thus output noise power,
P_n,out = ∫_(−W)^(W) (N₀/4) df = N₀W/2.

Therefore,
SNR_out = P_s,out/P_n,out = P₀W/(N₀W/2) = 2P₀/N₀.

SNR_out = 2P₀/N₀.

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