SSC JE Electrical Engineering 22 Jan 2018 Evening Paper

Question : 135 of 200

Marks: +1, -0

What is the value of current (in A) at half power of a series RLC cicuit at resonance, when the maximum value of current is 20 A?

Solution:

Here

P_1 = P

and

P_2 = P/ 2

Now power

P = I^2R

i.e.

P ∝ I_2

∴P_1/P_2=(I_1/I_2)^2

⇒2=400/I_2^2⇒I_2=10√2\A

= 14.14 A

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