This set of High Voltage Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Waveshape Control”.

1. An impulse voltage generator has a generator capacitance C_{1} of 0.02 μF, load capacitance C_{2} of 0.001 μF, front resistance of R_{1}=400 Ω and tail resistance of R_{2}=2980 Ω. The tail time is __________

a) 49.68 μs

b) 54.26 μs

c) 30.92 μs

d) 76.54 μs

View Answer

Explanation: Tail time is the discharging time. The capacitances C

_{1}and C

_{2}may be considered to be in parallel and discharging occurs through R

_{1}and R

_{2}. The tailing time is given by the equation

t=.7 (R

_{1}+R

_{2})(C

_{1}+C

_{2})

t=.7(400+2980) (.02+0.001) x 10

^{-6}

t=49.68 μs.

2. An impulse voltage generator has a generator capacitance C_{1} of 0.02 μF, load capacitance C_{2} of 0.001 μF, front resistance of R_{1}=400 Ω and tail resistance of R_{2}=2980 Ω. The charging time is __________

a) 1.9 μs

b) 1.14 μs

c) 2.0 μs

d) 0.5 μs

View Answer

Explanation: The time taken for charging is approximately three times the time constant of the circuit.

T=3.0 R

_{1}\(\frac{C_1 C_2}{C_1 + C_2}\)

T=3.0 x 400 x \(\frac{.02 * .001}{0.02+0.001}\) x 10

^{-6}

T=1.14 μs.

3. Wave shape control is flexible and independent.

a) True

b) False

View Answer

Explanation: Wave shape control is flexible and independent. Wave shape gets changed with the change in test object. But, the circuit is simple and hence is advantageous.

4. The transient voltage may be __________

a) an oscillatory wave

b) a damped oscillatory wave

c) critically damped wave

d) an oscillatory wave or a damped oscillatory wave

View Answer

Explanation: The transient voltage may be an oscillatory wave or a damped oscillatory wave. The frequency of the waves ranges from few hundred hertz to few kilo hertz. Transient voltage can also be considered as a slow rising impulse.

5. What is the time constant of the wave shaping circuit?

a) \(t=R_1 \frac{C_1 C_2}{C_1 + C_2}\)

b) \(t=2 R_1 \frac{C_1 C_2}{C_1 + C_2}\)

c) \(t=3R_1 \frac{C_1 C_2}{C_1 + C_2}\)

d) \(t=R_1 \frac{C_1 C_2}{2C_1 + C_2}\)

View Answer

Explanation: The time constant of the wave shaping circuit is \(=R_1 \frac{C_1 C_2}{C_1 + C_2}\). R

_{1}is the resistance, C

_{1}is the generator capacitance and C

_{2}is the load capacitance. The circuit inductance is very small and hence is not taken into account.

6. What is the time taken for charging the capacitance in the wave shape control circuit?

a) \(t=R_1 \frac{C_1 C_2}{C_1 + C_2}\)

b) \(t=2 R_1 \frac{C_1 C_2}{C_1 + C_2}\)

c) \(t=3R_1 \frac{C_1 C_2}{C_1 + C_2}\)

d) \(t=R_1 \frac{C_1 C_2}{2C_1 + C_2}\)

View Answer

Explanation: The time taken for charging the capacitance in the wave shape control circuit is \(t=3R_1 \frac{C_1 C_2}{C_1 + C_2}\). R

_{1}is the resistance, C

_{1}is the generator capacitance and C

_{2}is the load capacitance. The time taken for charging is thrice the time constant of the circuit. The circuit inductance is very small and hence is not taken into account.

7. What is the effective capacitance in wave shaping circuit?

a) \(C=\frac{C_1 C_2}{C_1 + C_2}\)

b) \(C=2 \frac{C_1 C_2}{C_1 + C_2}\)

c) \(C=\frac{C_1 C_2}{2C_1 + 2C_2}\)

d) \(C=3 \frac{C_1 C_2}{C_1 + C_2}\)

View Answer

Explanation: The effective capacitance in wave shaping circuit is \(C=\frac{C_1 C_2}{C_1 + C_2}\), C

_{1}is the generator capacitance and C

_{2}is the load capacitance. The circuit inductance is very small and hence is not taken into account.

8. What is the time taken for discharging in the wave shape control circuit?

a) \(t=(R_1+R_2) (C_1 + C_2)\)

b) \(t=.7(R_1+R_2) (C_1 + C_2)\)

c) \(t=3R_1 \frac{C_1 C_2}{C_1 + C_2}\)

d) \(t=R_1 \frac{C_1 C_2}{C_1 + C_2}\)

View Answer

Explanation: The time taken for discharging in the wave shape control circuit is \(t=.7(R_1 + R_2)(C_1 + C_2)\). It is also known as the tail time of the circuit. R

_{1}and R

_{2}is the resistance, C

_{1}is the generator capacitance and C

_{2}is the load capacitance. The time taken for charging is thrice the time constant of the circuit. The circuit inductance is very small and hence is not taken into account.

9. The generator capacitance is dependent on the design of the generator.

a) True

b) False

View Answer

Explanation: The generator capacitance is dependent on the design of the generator. The load capacitance is also fixed for a given design of the generator and test object. So to obtain the desired wave shape only the resistance can be controlled.

10. An impulse voltage generator has a load capacitance C_{2} of 0.001 μF, front resistance of R_{1}=400 Ω, tail resistance of R_{2}=2980 Ω and tail time 49.68 μs.The generator capacitance C_{1} of is __________

a) 0.001 μF

b) 0.2 μF

c) 0.02 μF

d) 0.03 μF

View Answer

Explanation: Tail time is the discharging time. The capacitances C

_{1}and C

_{2}may be considered to be in parallel and discharging occurs through R

_{1}and R

_{2}. The tailing time is given by the equation \(t=.7 (R_1 + R_2)(C_1 + C_2)\)

Hence, the generator capacitance C

_{1}can be found out by the following steps.

49.68 = \(.7(400+2980)(C_1 + 0.001)\)

C

_{1}= 0.02 μF.

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