This set of Power Electronics Aptitude Test focuses on “Single-Phase FW AC-DC-4”.
1. An SCR has the peak forward voltage = 1000 V. Find the maximum voltage that the SCR can handle if employed in a B-2 type full controlled converter circuit. Use factor of safety (FOS) = 2.5
a) 500 V
b) 400 V
c) 200 V
d) 1000 V
Explanation: In M-2 type configuration maximum voltage handled is Vm
Therefore, 1000/2.5 = 400 V.
2. A single-phase full converter bridge, is connected to a RLE load. The source has a rms voltage of 230 V and the average load current is 10 A. Find the firing angle for which the power flows from AC source to the DC load. Consider E = 120 V, R = 0.4 Ω.
Explanation: We have,
Vo = E + IR
Vo = 2Vm/π x cosα
Substitute the given values to discover α = 53.208°.
3. A single-phase full converter bridge, is connected to a RLE load. The source has a rms voltage of 230 V and the average load current is 10 A. Find the firing angle for which the power flows from the DC load to the AC source. Consider E = 120 V, R = 0.4 Ω, L = 2 Henry.
Explanation: We have,
Vo = E + IR
Vo = 2Vm/π x cosα . . . (i)
Substitute the given values to discover α = 124°
Note that as the power is flowing from DC to AC E has to be negative.
E = -120V.
Use 1 to find the firing angle.
4. For the below shown converter configuration, find the expression for the average value of voltage across the resister R with Vs = Vm sinωt and firing angle = α.
a) Vm/π cosα
b) 2Vm/π cosα
c) Vm/π (1+cosα)
d) 2Vm/π (1+cosα)
Explanation: Vo = Vo = 1/π x [∫ Vm sinωt d(ωt)] where the integration would run from α to π.
5. A single phase full controlled bridge converter is connected to load having R = 0.4 Ω, E = 120, L = 0.2 mH, Vs = 230V. The RMS value of load current is 10 A for a firing angle of 53.21°. Find the input power factor.
a) 0.5043 lag
b) 0.5391 lag
c) 0.6120 lag
Explanation: Io = Irms = 10A
Vs x Irms x cosɸ = E Io + Irms2 R.
6. A single phase full converter feeds power to a RLE load with R = 6 Ω and E = 60 V. Find the average value of the load current when the supply voltage is 230 V rms AC and a firing angle = 50°.
c) 12.1 A
d) 6.05 A
Explanation: Vo = 2Vm/π cosα = 133.084 V
I = Vo – E/R = 133.084-60 / 6 = 12.181 A.
7. A single phase full wave mid-point SCR converter, uses a 230/200 V transformer with the centre tap on the secondary side. The P.I.V. per SCR is
a) 100 V
Explanation: PIV for M-2 is Vm = √2Vs = √2 x 200.
8. In a single phase full converter, for discontinuous load current and extinction angle β > π, each SCR conducts for
Explanation: Each device would conduct for β-α.
9. A single-phase two pulse converter feeds RL load with a sufficient smoothing such that the load current does not fall to zero. If the resistance of the load circuit is increased then the
a) ripple content is not affected
b) ripple content of current increases
c) ripple content of current decreases
d) load current may fall to zero
Explanation: If the resistance of the load circuit is increased then the ripple content of current increases.
10. In case of controlled rectifiers, the nature of the load current (continues or discontinuous) depends upon the
a) type of load and firing angle
b) only on the type of load
c) only on the firing angle
d) it is independent of all the parameters
Explanation: It depends on both as firing angle will decide how fast and how much current flows. The load R, Rl or RLE can also effect the current depending upon the values of L and E.
Sanfoundry Global Education & Learning Series – Power Electronics.
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