Provided that the sinewave source is DC blocking and of low output impedance, you have the 1VAC (centered) riding on 1VDC.
Why not simulate it in the free LTSpice? E
PS: there is never (correction: seldom) such a thing as a stupid question
Point A is the measurement across the AC voltage, so the AC source will determine what the voltage will be. The DC source only determines the amount, direction, and type of current existing in the circuit.
Assume your AC signal is goes from -1V to +1V (i.e.it has no DC component itself - DC blocking) then when the sine wave is at -1V, it is added to the +1V of the DC source which gives you 0V. When the sine wave is at +1V and added to the 1V DC source, the output is 2V, so as canadaelk has aleady pointed out, it'a a 1V sinewave sat on a 1V DC level. What used to vary from -1V to +1V now varies from 0C to +2V.
but point A is also across DC source in series to resistor.
so the AC source will determine what the voltage will be. The DC source only determines the amount, direction, and type of current existing in the circuit.
Assume your AC signal is goes from -1V to +1V (i.e.it has no DC component itself - DC blocking) then when the sine wave is at -1V, it is added to the +1V of the DC source which gives you 0V. When the sine wave is at +1V and added to the 1V DC source, the output is 2V, so as canadaelk has aleady pointed out, it'a a 1V sinewave sat on a 1V DC level. What used to vary from -1V to +1V now varies from 0C to +2V.
We don't know enough about the signal source to determine the output voltage of this circuit.
Is it a signal generator with an average DC output of 0V but an output impedance of 50 ohms?
Is the signal from an amplifier that has an extremely low output impedance and an average output of 0V?
Or does the signal generator have an output coupling capacitor?
Does the signal generator have a high output impedance so that the 50 ohms load here shorts its signal?
Can you please explain for each case? ie
if signal generator has average DC output of 0V but an output impedance of 50 ohms
if the signal from an amplifier has an extremely low output impedance and an average output of 0V
if signal generator have an output coupling capacitor
if the signal generator have a high output impedance so that the 50 ohms load here shorts its signal
Please Sir..... This is for understanding purpose
Then the 50 ohms of the signal generator and the 50 ohms from the power supply form a voltage divider and reduce the AC signal and the average DC output to 0.5V. The output will be a 0.5V signal riding on +0.5V so it will swing from 0V to +0.5V.
if the signal from an amplifier has an extremely low output impedance and an average output of 0V
You said in your previous post that the AC voltage source was "ideal". And since the AC source is grounded at the voltage reference point, point A is going to be the same as the output from the AC source, regardless of the value of the DC source.