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# Test: Power Electronics - 2

## 10 Questions MCQ Test RRB JE for Electrical Engineering | Test: Power Electronics - 2

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This mock test of Test: Power Electronics - 2 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Power Electronics - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Power Electronics - 2 quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Power Electronics - 2 exercise for a better result in the exam. You can find other Test: Power Electronics - 2 extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

### In a thyristor

Solution:

Latching current is the minimum value of anode current which it must attain during turn – ON process to maintain conduction when gate signal is removed.

Holding current is the minimum value of anode current below which it must fall for turning – OFF the thyristor.

QUESTION: 2

### Read the given statements below regarding turning – on an SCR with a gate signal. (1)  An SCR must be forward – biased. (2)  Gate pulse width must be less than the turn-on time of an SCR. (3)  Anode to cathode voltage must be more than finger voltage. (4)  The magnitude of gate current must be less than the minimum gate current required to turn – on a thyristor. (5)  The magnitude of gate current must be less than the maximum gate current allowed  Which of the above statements are false.

Solution:

1. Anode must be positive with respect to cathode.

2.  Gate pulse must be more than the turn – on time of an SCR. This will ensure that anode current exceeds the latching current before gate signal is removed.

3.  A finger voltage is that voltage below which an SCR cannot be turned on with a gate signal.

4.  Magnitude of gate current must be more than the minimum gate current required to turn – on a thyristor.

QUESTION: 3

### An SCR is rated for 700 V PIV. What is the value for which the device can be operated if the voltage safety is 2.

Solution:

P.I.V = Vm × Voltage safety factor

700 = Vm × 2

Vm = 350 V QUESTION: 4

What should be the input voltage to turn off the SCR in the circuit shown in figure below when holding current of the SCR is 4 mA and load resistance is 4 kΩ. Solution:

To turn off the SCR anode current should go less than holding current

⇒ V = 4 × 10-3 × 4 × 103

⇒ V = 16 V

Form the options, V = 15 V is correct. for the remaining options, holding current is less than anode current.

QUESTION: 5

SCR with a rating of 1200 V and 250 A are available to be used in a string to handle 8 kV and 1kA. The number of parallel and series units required are respectively _____ (Take derating factor is 0.2)

Solution:

Derating factor = 1 – string efficiency QUESTION: 6

An SCR has half cycle source current rating of 3000 A for 50 Hz supply. It’s one cycle surge current and I2t rating respectively are

Solution:

Let I and Isb be the one – cycle and sub – cycle surge current ratings of the SCR respectively t = 10 ms, T = 20 ms ⇒ I = 2121.32 A

I2t rating = I2 × 1/2f = 45,000 Amp2.sec

*Answer can only contain numeric values
QUESTION: 7

For an SCR, the gate – cathode characteristic has a straight line slope of 150. For trigger source voltage of 20 V and allowable gate power dissipation of 1.2 watts, the gate – source resistance is _______ (in Ω)

Solution:

VgIg = 1.2 W *Answer can only contain numeric values
QUESTION: 8

Thyristor shown in figure below has I2t rating of 60 A2s. If terminal A gets short circuited to ground, calculate the fault clearance time so that SCR ins not damaged – (in ms) Solution:

Maximum fault current occurs when source voltage is at its peak = 240 √2 V

When terminal ‘A’ gets short – circuited to ground the resistance offered to the source

= 10+(20||20) = 20 Ω ⇒ tc = 208.33 ms

*Answer can only contain numeric values
QUESTION: 9

It is required to operate 200 A SCR in parallel with 300 A SCR with their respective ON – State voltage drop of 2 V and 1.2 V. calculate the value of resistance to be inserted in series with each SCR so that they share the total load of 500 A in proportional to their current ratings – (in mΩ)

Solution:

Dynamic resistance of SCR with 200 A

= 2/200 = 0.01Ω = 10 mΩ
Dynamic resistance of SCR with 300 A

= 1.2/300 = 0.004Ω = 4 mΩ
Let Rs be the resistance inserted in series with each SCR.

Current shared by SCR= 500 Current shared by SCR= 500  ⇒ 12 + 3 Rs = 20 + 2Rs

⇒ Rs = 8 m Ω

QUESTION: 10

Four 800 V thyristors are connected in series to share 3 kV supply voltage. The maximum forward leakage current is 4 mA, and the maximum difference in recovery charge is 20 μC. Determine the value of R1 and C (assuming equal voltage sharing).

Solution:

Here, n = 4 