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For the circuit, if capacitor is shorted,then what is amplitude of output voltage in mA?

A)0 B)25 C)50 D)75 C)100

 

 

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Asked by miaowmira1900 24th January 2020, 5:30 PM
Answered by Expert
Answer:
It is assumed point-C in the circuit is at ground potential. As shown in figure, after shorting the capacitor,
let us assume at junction B, current i1 is getting divided as i2 and i3.
 
For the line, A-B-C, if we write kirchoff's voltage law,   300(i1 + i3) = 5  ..................(1)
 
For the line A-B-E, neglecting 100 mV ac voltage source, we have,  300 i1 + 150 i2 + 4 = 5 ............(2)
 
At junction B,  we have ,  i1 = i2 + i3   ......................(3)
 
By solving above equations, we get i3 = 10.3 mA
 
Hence output voltage = 10.3 mA × 300 Ω = 3.09 V
Answered by Expert 25th January 2020, 9:53 AM
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