alternating current circuit help

[Bhuvanesh123]

i have attached my circuit and corresponding waveform,i don't even know what my attached circuit is called as.i have 5vrms ac source c, 3v dc source ,some k ohm resistor , all are connected in series.my output waveform look like its shifted down.what operation actually going on there and how the waveform shifted down.Thank you in advance

 

[D.A.(Tony)Stewart]

if 5Vpp then it will look like your pix.

but if 5Vrms then peak voltage is 5* sqrt(2)
Vo = 5 sqrt(2) sint (wt) -3V

follow KVL rules.

**broken link removed**

 

[Bhuvanesh123]

ohh ,kvl that is that magic word i missed,thanks .
Now, 3v dc source is replaced with forward biased pn diode

Vo = 5 sqrt(2) sint (wt) -0.7V

am i right?

 

[arouse1973]

You have a D.C source which maintains a fixed potential difference between its two terminals. Because the output is taken from the lower potential side of the battery then this point will always be 3 Volts lower than the positive potential terminal of the battery. So when your sine wave (we will say it's peak to peak 10 volts) goes to the maximum positive value of 5 volts then the output must be 3 Volts lower. So the maximum output will be 2 Volts. When the sine wave goes negative then the output terminal will be 3 Volts lower still and produce -8 Volts. This is what causes the waveform to shift down by 3 Volts. This is referenced to the lower end of the A.C source.
Thanks
Adam