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# Test: Single Phase HW AC-DC - 1

## 20 Questions MCQ Test Power Electronics | Test: Single Phase HW AC-DC - 1

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This mock test of Test: Single Phase HW AC-DC - 1 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 20 Multiple Choice Questions for Electrical Engineering (EE) Test: Single Phase HW AC-DC - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Single Phase HW AC-DC - 1 quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Single Phase HW AC-DC - 1 exercise for a better result in the exam. You can find other Test: Single Phase HW AC-DC - 1 extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

### A voltage source Vs = Vm sinωt is connected in series with a resistance R and an SCR. At some firing angle delay of α a positive gate pulse is applied to the SCR which turns on the SCR. Considering ideal conditions, at the instant α the voltage at the resistor terminals Vo

Solution:

At the firing angle, the SCR turns on and thus the resistor or load R is connected to the source.

QUESTION: 2

### The average output voltage is maximum when SCR is triggered at ωt =

Solution:

The sooner the conduction starts higher the average power. Though practically a device cannot be triggered at exactly zero degrees.

QUESTION: 3

### In the below given circuit, the SCR is triggered at a certain firing angle α. The load current falls to zero at ωt =

Solution:

The SCR is naturally commutated at 90°.

QUESTION: 4

In the method of phase control, the phase relationship between ___ & ___ is controlled by varying the firing angle

Solution:

By varying the firing angle, the load current and load voltage can be controlled.

QUESTION: 5

In a single phase half-wave thyristor circuit with R load & Vs=Vm sinωt, the maximum value of the load current can be given by

Solution:

Vm is the peak value of the load as well as supply voltage. I = Vm/R.

QUESTION: 6

Below shown ideal circuit is triggered at a firing angle of α. The voltage across the SCR is zero from ωt = Solution:

Before the SCR is triggered all of the supply voltage appears across it that is from 0 to α, at α the device is gated and it starts to conduct acting like a S.C (ideally). Hence voltage across it is zero from α to π. At π the SCR is again naturally commutated due to the reversal of the AC supply & a negative voltage appears across it.

QUESTION: 7

Below shown ideal circuit is triggered at a firing angle of α. From ωt = π to 2π+α Solution:

From ωt = π to 2π+α, negative cycle is active and the SCR experiences revered biased voltage by Vs.

QUESTION: 8

For a single phase thyristor circuit with R load & firing angle α, the conduction angle can be given by

Solution:

SCR is fired at α, hence it conducts from α till it is commutated by the reversal of the AC supply i.e at π radians.

QUESTION: 9

For the below given circuit with firing angle α & angular frequency of ω, the circuit turn-off time in secs can be given by Solution:

The circuit turns of at ωt = π as it is a half wave circuit.
Hence toff = π/ω.

QUESTION: 10

A single-phase half-wave thyristor circuit with R load is triggered at an angle of α = 0°. As such, the maximum value of the average output voltage would be given byConsider Vs = Vm sinωt.

Solution:

For the whole cycle of 2π, the R load would be active only for half of the cycle. Hence, Vo = Vm/π.

QUESTION: 11

For a certain SCR configuration, the below shown waveform is obtained. Find the value of the average output voltage with α = 30° & Vs = 240 V. Solution:

Vm = √2 Vs = 339.41 V
Vavg = ∫ Vm Sin ωt d(ωt) where the integration runs from α to π
Vavg = Vm/2π (1+cos α).

QUESTION: 12

A single-phase half wave circuit has Vs = 230 V with a R load of 100 Ω. Find the average load current at α = 30°.

Solution:

Io = Vo/R
Vo = (Vm/2π) x (1+cosα).

QUESTION: 13

A single phase 230 V, 1 kW heater is connected across a 1-phase 230 V, 50 Hz supply through an SCR. The firing angle of the SCR is adjusted to give a rms voltage of 155 V. Find the power absorbed by the heater element.

Solution:

R (Heater element resistance) = 2302/1kW = 52.9 Ω
Now, P = Vrms2/R = 1552/52.9.

QUESTION: 14

For a single phase half-wave thyristor circuit with R load, the power delivered to the resistive load is

Solution:

P = I2.R
V = IR
I = V/R
Hence, P = V2/R.

QUESTION: 15

For a single phase half-wave thyristor circuit with R load, the input power factor is given by

Solution:

(p.f)input = power delivered to load/input VA.

QUESTION: 16

For the below shown circuit, The load voltage waveform from ωt = π to ωt = π+α = β (extinction angle) is

Solution:

In this case, the Inductor L is force conducting the SCR & supply current to the source the current is gradually decaying as the inductor is discharging.

QUESTION: 17

For the below shown circuit, The SCR is gated at α. At ωt = π

Solution:

At π the L force conducts the SCR & current starts gradually decaying whereas at π the voltage is zero as it goes from positive to negative at π.

QUESTION: 18

In case of a single-phase half-wave circuit with RL load, with firing angle α and extinction angle β, the conduction angle γ can be written as

Solution:

Extinction angle is some angle > π at which load current reduces to zero. It depends upon the inductor value.

QUESTION: 19

For a half-wave single phase thyristor rectifier circuit with any type of load the _______ + _________ waves must ideally give the supply voltage waveform. Solution:

The supply voltage either appears across at the thyristor or the load.

QUESTION: 20

In case of a single-phase half-wave circuit with RL load, with firing angle α and extinction angle β, the thyristor is reversed biased from

Solution:

As the load current goes to zero from β, the SCR is reversed biased or in the OFF state till it is turned on at 2π again by the positive AC cycle.