# High Voltage Engineering Questions and Answers – Waveshape Control

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

1. An impulse voltage generator has a generator capacitance C1 of 0.02 μF, load capacitance C2 of 0.001 μF, front resistance of R1=400 Ω and tail resistance of R2=2980 Ω. The tail time is __________
a) 49.68 μs
b) 54.26 μs
c) 30.92 μs
d) 76.54 μs

Explanation: Tail time is the discharging time. The capacitances C1 and C2 may be considered to be in parallel and discharging occurs through R1 and R2. The tailing time is given by the equation
t=.7 (R1+R2)(C1+C2)
t=.7(400+2980) (.02+0.001) x 10-6
t=49.68 μs.

2. An impulse voltage generator has a generator capacitance C1 of 0.02 μF, load capacitance C2 of 0.001 μF, front resistance of R1=400 Ω and tail resistance of R2=2980 Ω. The charging time is __________
a) 1.9 μs
b) 1.14 μs
c) 2.0 μs
d) 0.5 μs

Explanation: The time taken for charging is approximately three times the time constant of the circuit.
T=3.0 R1 $$\frac{C_1 C_2}{C_1 + C_2}$$
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

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

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

Explanation: The time constant of the wave shaping circuit is $$=R_1 \frac{C_1 C_2}{C_1 + C_2}$$. R1 is the resistance, C1 is the generator capacitance and C2 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) $$t=R_1 \frac{C_1 C_2}{2C_1 + C_2}$$

Explanation: The time taken for charging the capacitance in the wave shape control circuit is $$t=3R_1 \frac{C_1 C_2}{C_1 + C_2}$$. R1 is the resistance, C1 is the generator capacitance and C2 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) $$C=3 \frac{C_1 C_2}{C_1 + C_2}$$

Explanation: The effective capacitance in wave shaping circuit is $$C=\frac{C_1 C_2}{C_1 + C_2}$$, C1 is the generator capacitance and C2 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) $$t=R_1 \frac{C_1 C_2}{C_1 + C_2}$$

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. R1 and R2 is the resistance, C1 is the generator capacitance and C2 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

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 C2 of 0.001 μF, front resistance of R1=400 Ω, tail resistance of R2=2980 Ω and tail time 49.68 μs.The generator capacitance C1 of is __________
a) 0.001 μF
b) 0.2 μF
c) 0.02 μF
d) 0.03 μF

Explanation: Tail time is the discharging time. The capacitances C1 and C2 may be considered to be in parallel and discharging occurs through R1 and R2. The tailing time is given by the equation $$t=.7 (R_1 + R_2)(C_1 + C_2)$$
Hence, the generator capacitance C1 can be found out by the following steps.
49.68 = $$.7(400+2980)(C_1 + 0.001)$$
C1 = 0.02 μF.

Sanfoundry Global Education & Learning Series – High Voltage Engineering.

To practice all areas of High Voltage Engineering, here is complete set of Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}