SSC JE Electrical Engineering 2014 Set 1 Paper

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Question : 134 of 200

Marks: +1, -0

Calculate the total susceptance of the circuit shown in figure.

6.67 mho
1.87 mho
0.16 mho
0.08 mho

Solution:

The susceptance of given circuit will be

B_1 + B_2

here

B_1 = X_L∣(R^2_L+ X ^2_L)

4|( 3 ^2+4^2)= 0.16 (\mho)

B_2 = X_C∣(R^2_C+ X ^2_C)

- 10|( 5^2+ ( -10) ^2)= -0.08 \mho

Susceptance

= 0.16 -0.08 = 0.08

mho

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