This set of Electromagnetic Theory Assessment Questions and Answers focuses on “Smith Chart”.

1. The Smith chart is a polar chart which plots

a) R vs Z

b) R vs Znorm

c) T vs Z

d) T vs Znorm

View Answer

Explanation: The Smith chart is a frequency domain plot. It is the polar chart of the reflection coefficient R with respect to the normalised impedance Znorm.

2. The Smith chart is graphical technique used in the scenario of transmission lines. State true/false.

a) True

b) False

View Answer

Explanation: The Smith chart is used for calculating the reflection coefficient and standing wave ratio for normalised load impedance of a transmission line.

3. The Smith chart consists of the

a) Constant R and variable X circles

b) Variable R and constant X circles

c) Constant R and constant X circles

d) Variable R and variable X circles

View Answer

Explanation: The Smith chart consists of the constant resistance circles and the constant reactance circles. The impedances are plotted using these circles. Also stub matching can be done using the Smith chart.

4. The circles in the Smith chart pass through which point?

a) (0,1)

b) (0,-1)

c) (-1,0)

d) (1,0)

View Answer

Explanation: All the constant resistance and reactance circles in the Smith chart pass through the (1,0) point. This is the midpoint of the Smith Chart. The resistance is unity and reactance is zero at this point.

5. Moving towards the clockwise direction in the Smith chart implies moving

a) Towards generator

b) Towards load

c) Towards stub

d) Towards waveguide

View Answer

Explanation: On moving towards the clockwise direction in the Smith chart, we are traversing towards the generator. This is used to calculate the normalised load impedance.

6. The centre of the point having a normalised resistance of 1.2 ohm and reactance of 1.5 ohm is

a) (0.54,0)

b) (0.45,0)

c) (0.36,0)

d) (0.78,0)

View Answer

Explanation: The centre of a point in Smith chart is given by C = (r/1+r, 0). On substituting for r = 1.2, we get centre as (1.2/1+1.2,0) = (0.54,0).

7. The normalised load impedance of the transmission line 50 ohm with a load of 30 ohm is

a) 30

b) 150

c) 5/3

d) 3/5

View Answer

Explanation: The normalised impedance is calculated by dividing the impedance with the characteristic impedance. Given that the load impedance is 30 ohm, the normalised load impedance of the 50 ohm transmission line is 30/50 = 3/5 ohm.

8. The radius of the point having a normalised resistance of 1 ohm is

a) 1

b) 0.2

c) 0.5

d) 0.25

View Answer

Explanation: The radius of the point with a radius r is given by R = 1/r+1. On substituting for r = 1, we get R = 1/1 + 1 = ½ = 0.5.

9. The best stub selection for the transmission line will be

a) Series open

b) Series short

c) Shunt open

d) Shunt short

View Answer

Explanation: Normally series stubs are not preferred as modification of the stub parameters requires changing the whole stub setup. Shunt stubs enable modification with ease. Open circuited stubs are not preferred as it will radiate power like an antenna, which is undesirable. Hence shorted stubs are used.

10. The centre and radius of a line with normalised load impedance of 1 + 0.5j is

a) (1,2) and 2

b) (2,1) and 2

c) (1,2) and 1

d) (2,1) and 1

View Answer

Explanation: The centre and radius of a line are (1, 1/x) and 1/x, where x is the reactance. Here x = 0.5, from the given data. Thus C = (1,2) and R = 2.

**Sanfoundry Global Education & Learning Series – Electromagnetic Theory.**

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