Question

(a) By means of a schematic circuit diagram illustrate the peak rectifier. If the supply voltage is v(t) = Vm Sin wt. what is the voltage across the load resistor? (6)

(b) A d.c. motor takes an armature current of 110 A at 480 V. The resistance of the armature circuit is 0.2 Ω. The machine has 6 poles and the armature is lap connected with 864 conductors. The flux per pole is 0.05 Wb. Calculate: (10)

(a) The speed,

(b) The gross torque developed by the armature.

Answer 1

Part (a)

Peak rectifier:

A peak rectifier (peak detector) consists of a diode in series with a capacitor and a load resistor. The diode rectifies the a.c. input and charges the capacitor to the peak value of the input voltage.
Circuit: a.c. source -> diode -> node (capacitor to earth, load resistor to earth). Output taken across the capacitor/load.
During the positive half cycle the diode conducts and charges the capacitor to the peak value Vm. When the input falls below the capacitor voltage the diode becomes reverse biased and stops conducting; the capacitor discharges slowly through the load. If the time constant R x C is large compared with the period, the output is held near Vm.
If the supply voltage is v(t) = Vm sin(wt), the voltage across the load resistor is approximately the peak value Vm (for an ideal diode and large time constant), i.e. a d.c. voltage close to Vm with small ripple.

Part (b)

D.C. motor: armature current 110 A at 480 V, armature circuit resistance 0.2 ohm, 6 poles, lap-connected armature with 864 conductors, flux per pole 0.05 Wb.

Back e.m.f. E = V - Ia Ra = 480 - 110 x 0.2 = 480 - 22 = 458 V.
For a lap-connected armature, number of parallel paths A = number of poles P = 6.
E.m.f. equation: E = (P x Z x phi x N) / (60 x A). Since A = P, E = (Z x phi x N)/60.
(i) Speed: N = (E x 60)/(Z x phi) = (458 x 60)/(864 x 0.05) = 27480/43.2 = 636.1 rev/min.
(ii) Gross torque developed: T = (P x Z x phi)/(2 pi A) x Ia = (6 x 864 x 0.05)/(2 x 3.1416 x 6) x 110 = (259.2/37.70) x 110 = 6.876 x 110 = 756.4 N m.
(Check: armature power = E x Ia = 458 x 110 = 50,380 W; angular speed = 2 pi x 636.1/60 = 66.6 rad/s; T = 50380/66.6 = 756.5 N m.)

So speed = 636 rev/min and gross torque = 756 N m.

✓ Verified Model Answer (Text Solution)

Peak rectifier:

D.C. motor: armature current 110 A at 480 V, armature circuit resistance 0.2 ohm, 6 poles, lap-connected armature with 864 conductors, flux per pole 0.05 Wb.