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Draw the output waveform across the resistor. |
Answer» Solution : Here, current passing through the closed path given in the STATEMENT, `I=(v_(i))/(R+R_(D))""…(1)` (where `R_(D)=` RESISTANCE of diode) During negative HALF cycle of input voltage, SINCE diode is reverse biased, `R_(D)=OO` and so from equation (1), I = 0 and so output voltage `v_(0)=IR=0 ""(because I=0)` During positive half cycle of input voltage, since diode is forward biased , `R_(D)=0` and so from equation (1), `I=(v_(i))/(R )` and so output voltage `v_(0)=IR =((v_(i))/(R ))R=v_(i)=1`volt. (from figure) |
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