1.

A 50 μF capacitor, when connected in series with a coil having resistance of 40Ω, resonates at 1000 Hz. The circuit is in resonating condition. The voltage across the coil is __________(a) 100.31 V(b) 200.31 V(c) 300.31 V(d) 400.31 VThe question was asked in examination.This interesting question is from Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory

Answer»

The correct choice is (a) 100.31 V

To explain: At resonance, | I | = \(\frac{V}{Z} = \frac{V}{R} \)

Given that, R = 40Ω and V = 100 V

∴ | I | = \(\frac{100}{40}\) = 2.5 A

XL = ωL = 2π × 1000 × 0.5 × 10^-3

= 2π × 0.5 = 3.14 Ω

∴ VCOIL = I Z (At resonance)

= 2.5 \(\SQRT{40^2 + 3.14^2}\)

= 2.5 × 40.122 = 100.31 V.



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