1 what is the time domain expression of this voltage 2 find

1. what is the time domain expression of this voltage?

2. find the time domain expression when another voltage with similar amplitude and frequency lags the current voltage by 30?

. Tte follawing wwetonn was recorded by connecting as uesxcilloscope proke to the wall cutlet. Answer the tollowin l questions based on this wavefureme time ims)

Solution

A normal voltage wave is denoted by the expression V(t) = Vm Sin(t±) where Vm signifies the peak value of voltage , the angular frequency and , the phase difference .

Mathematically =2**f

where f is the frequency of the wave.

Coming back to your question answer to Ques-1 is 165 Sin 392500t

                                                          answer to Ques-2 is 165 Sin (392500t-30)

explanation

ques 1

Here the waveform have a peak value of 165 volt

The time period of the signal(the time to complete on ac cycle) is .016 ms. Therefore the frequency is 1/(.016*10-3)

By substituting these values on =2**f we get =2*3.14*1/(.016*10-3)

                                                                                                   =392500

So expression for voltage is 165 Sin 392500t.

ques 2

Since the second voltage lags the first voltage by 30 degree the should be given as negative because the lagging signal should have the peak value only after 30 degrees of first voltage attaining the peak value in a scale of angle on x axis. x(t-30) only causes shifting of signal to 30 units more to the right of original signal thus fulfilling our criteria

A normal voltage wave is denoted by the expression V(t) = Vm Sin(t±) where Vm signifies the peak value of voltage , the angular frequency and , the phase difference .

Mathematically =2**f

where f is the frequency of the wave.

Coming back to your question answer to Ques-1 is 165 Sin 392500t

                                                          answer to Ques-2 is 165 Sin (392500t-30)

explanation

ques 1

Here the waveform have a peak value of 165 volt

The time period of the signal(the time to complete on ac cycle) is .016 ms. Therefore the frequency is 1/(.016*10-3)

By substituting these values on =2**f we get =2*3.14*1/(.016*10-3)

                                                                                                   =392500

So expression for voltage is 165 Sin 392500t.

ques 2

Since the second voltage lags the first voltage by 30 degree the should be given as negative because the lagging signal should have the peak value only after 30 degrees of first voltage attaining the peak value in a scale of angle on x axis. x(t-30) only causes shifting of signal to 30 units more to the right of original signal thus fulfilling our criteria