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GATE 2015 EE - SET 2 - Question 8

Solution:

100\piRC = 50

R= \frac{50}{100\pi c}

During positive half cycle

V_{0}  = \frac{V_{in}(R)}{(R-jX_{c})} , where R = 50X_{c}

V_{0} = \frac{v_{in}R}{R-jX_{c}} = \frac{v_{in}(50X_{c})}{-jX_{c}}

V_{0} = \frac{100sin(100\pi t)50X_{c}}{50X_{c}-jX_{c}}

Positive half cycle \Rightarrow V_{0}(peak) = \left | 100 \right |

Negative half cycle \Rightarrow V_{0}(peak) = \left | -100 \right |

V_{0}\left | 100 \right |\left | -100 \right | = 200 V

 

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