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GATE 2015 EE - SET 2 - Question 45

Solution :

V_{P} = ± 15 V ,T = 0.4\pi ms ,R=5Ω , L = 10 * 10^{-3}H , C + 4 * 10^{-6}F , w = 5000 rad/sec , (V_{c})_{P} = ?

Fourier series of a square wave ,

\sum_{n= 1,3,5}^{\infty } \frac{4V}{n\pi } sin\omega t

\sum_{n= 1,3,5}^{\infty } \frac{4(15)}{n\pi } sin\omega t

\sum_{n= 1,3,5}^{\infty } \frac{19.10}{n\pi } sin\omega t

X_{c} = \frac{1}{w{c}} = \frac{1}{5000*4*10^{-6}} = 50 Ω

X_{c} =WL = 5000 *10 * 10^{-3} = 50Ω

X_{c}  = X_{L} = ckt is in resonance

V_{c}= I_{fundamental}*X_{c}*Sin\omega t

V_{c} = \frac{V_{fundamental}}{R} * X_{c} Sin\omega t

V_{c} =\frac{19.10}{5} * 50 Sin\omega t
V_{c} = 191 Sin\omega t

Amplitude of V_{c} = 191V

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