KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 100 of 120

Marks: +1, -0

Consider the R-L-C circuit given below. The circuit is driven by a 50 Hz AC source with peak voltage 220 V. If R = 400 Ω, C = 200 µF and L = 6 H, the maximum current in the circuit is closest to [2017]

0.120 A
0.55 A
1.2 A
5.5 A

Validate

Solution:

X_L = ωL = 2π × 50 × 6 = 600 πΩ

X_C =1/{ωC}=1/{2π80×200×10^{-6}}

=100/{2π}=50/πΩ

I_\max =V/z=220/√{R^2+(X_L-X_C)^2

=220/√{400^2+(600π-50/π)^2

I_\max = 0.120 A

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