Question

An AC source of frequency 50 Hz is connected to 50 mH inductor and a bulb. To get the maximum brightness from the bulb, what should be the capacitance of a capacitor connected in series with the circuit?

• A. $ 50\times {{10}^{-4}}F $
• B. $ 1\times {{10}^{-4}}F $
• C. $ 2.03\times {{10}^{-4}}F $
• D. $ 3\times {{10}^{-4}}F $

Answer 1

Answer: (C)

For maximum brightness maximum current is required in the circuit. To get maximum current,
$ {{X}_{L}}={{X}_{C}} $ (at resonance)
$ \Rightarrow $ $ 2\pi \,vL=\frac{1}{2\pi vC} $
$ \Rightarrow $ $ C=\frac{1}{4{{\pi }^{2}}{{v}^{2}}L} $
$ =\frac{1}{4{{(3.14)}^{2}}{{(50)}^{2}}(50\times {{10}^{-3}})} $
$ =2.03\times {{10}^{-4}}F $