SSC JE Electrical Engineering 2013 Paper

Question : 182 of 200

Marks: +1, -0

A current from an ac source bifurcates into two branches A and B in parallel. Branch A is an inductor with 30 µ H inductance and 1 Ω resistance. Branch B is another inductor with inductance L and 1.5 Ω resistance. For the ratio of currents in the branches to be independent of supply frequency, value of L should be

30.5 µ H
20 µ H
45 µ H
29.5 µ H

Validate

Solution:

I=I_A+I_B

I_A/I_B={1.5+jWL}/{1+jW30}

I_A/I_B={(1.5+jWL)(1-W30µ)}/{1+(w30)^2}

={1.5+jWL-jW45µ+W^2L30µ}/{1+(30W)^2}

For independent of supply frequency
WL – W 45 µ = 0
L = 45 µ H

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