# AC Voltage Sources Riding A DC Voltage Source

1. Oct 5, 2013

### Physixs

Hello again, I am having a bit of trouble understanding superimposed voltage.

Here is the problem:
A sinusoidal AC voltage source is in series with a DC source. Effectively, the two voltages are superimposed. Draw the total voltage across RL. Determine the maximum current through RL and the avg across RL.

Then there is a diagram.
The diagram starts with:
ground --> 200V DC --> 150Vpp AC --> R1 (47Ohm) --> RL (100 ohm) --> ground

I understand that the AC signal will be "increased" to a new peak voltage of 350V.
I understand that the AC signal will have a new low or 50V

What I am not sure about which method below (if either) is correct

1. Since the AC voltage never dips below zero again (due to the DC source) would the RMS value of the voltage simply be the DC source (200V)? Or, would I still calculate the RMS using the 150Vp (350 minus the new equilibrium base of 200V = 150Vp)?

2. Or, should I calculate the Vrms of the AC signal then simply add that value to the DC signal to get the "total voltage" in the circuit (since they are in series)?

I am not sure of the order of operations here. Do I combine the voltages then calculate? Or calculate then combine? Or neither?

I am having a lot of problems with this because the second part of the questions says find the "average voltage dropped across RL"). This statement implies that the voltage will still be alternating (otherwise there would be no average). This confused me because the method I used, was the Vrms which, with a superimposed DC signal of the 200V magnitude, would make it quite similar to the Vrms being dropped? I think?

I am not even sure I can word my question correctly because I am so confused. Sorry.

I am lost on how to do this problem (it wasn't assigned in class). When I asked my TA, he wasn't sure either (it was one of the challenge problems). I wanted to do it on my own, however, since I am struggling with RMS voltage (not understanding the exact point of using it if there is a DC source voltage present that increases the AC signal above zero).

Any insight would be great. Thank you!

2. Oct 5, 2013

### UltrafastPED

If the only source was pure AC you would have a sinusoid which goes above and below the zero volt point symmetrically - you would see the same thing on your oscilloscope with AC coupling on or off.

But when you add in the DC voltage it shifts this center of symmetry by a fixed amount (eg, + or - 5 volts). Now your oscilloscope will show this shift if AC coupling is off, but will remove it (set it back to zero) if AC coupling is on.

You can simply solve the problem with resistors twice: first AC, then for the DC, then add them together.

3. Oct 6, 2013

### Physixs

Thank you for your response. I just wanted to clarify.

So I calculate the RMS of the AC wave first, for 150V that is106Vrms. Then I can calculate the voltage drop over RL (the 100 ohm resistor) which would be 72.15V

I then calculate the DC voltage drop which would be 136V

this is the part I am lost on. The answer in the back of the book says 136V is the average (which is the voltage produced by just the DC drop)

I guess what I am saying is that I do not understand (internally/electric-ly) what is going on to grant a voltage drop of 136V across that resistor. Does the AC signal just cancel itself out? Will it always be irrelevant if the DC signal is powerful enough to increase the AC signal's lowest point above zero?

I don't really understand why the AC signal doesn't matter.

4. Oct 6, 2013

### Drakkith

Staff Emeritus
To go out on a limb here:
It looks to me like the AC voltage doesn't matter because you never reverse polarity. If you had zero DC voltage, then half the time you would have current flowing the other way through the circuit, but with your DC voltage offset so far that the AC never takes it below zero, it appears to simply cancel itself out.

Normally we have to account for the current through the circuit both ways, as power is being transferred during both alternations. But in this setup, you are simply reducing the current flow during the negative alternation of the AC signal, which doesn't increase the power used during that time, it actually reduces it.

How's that sound?

5. Oct 6, 2013

### Physixs

Sounds great! Thank you that makes a lot of sense now!

6. Oct 6, 2013

### Drakkith

Staff Emeritus
Make sure to ask someone who actually knows the answer before you lock that down in your head. I really don't know, I've never dealt with this kind of problem before.

7. Oct 6, 2013

### Physixs

haha oh ok I will wait to see what someone else says. What you said seemed to make perfect sense though.

8. Oct 6, 2013

9. Oct 6, 2013

### Physixs

The article was helpful in understanding the derivation of offset voltage, however, when I apply the equation to this challenge problem:

VRMS = (200^2 + (150^2)/2)^.5 = 226.38V
current would then be 226.38/147 = 1.54 A
Drop across 100 Ohm would be 154VRMS

Solving for VP, I would get 154V√2 = 217.8Vp (the voltage drop in terms of Vp)
Plugging that into the avg voltage formula: Vavg = 2x217.8/π = 138.7V

The back of the book says it is 136V. Did I do something wrong?

10. Oct 6, 2013

### Staff: Mentor

This thread would probably have a better home in the Electrical Engineering forum, or better yet, the Engineering, Comp Sci, & Technology Homework forum. I'm debating whether to move it there. In the meantime...

The common expressions that are given to relate Vp to Vave to Vrms for pure sine waves do not apply to other waveforms. In such cases it's usually simpler and safer to go back to basics and invoke the definitions of those quantities.

Thus, if a waveform has a period T and is described by a function f(t), then
$$Average = \frac{1}{T}\int_0^T f(t) dt$$
which is clearly the average of value of the function over one period, and

$$RMS = \sqrt{\frac{1}{T}\int_0^T f(t)^2 dt}$$
The square Root of the Mean of the Square values of the function over one period.

When the varying part of the function is sinusoidal you use the angular form f(θ), and the angle for a period being $2 \pi$ radians.

For your problem start by "solving" the circuit to determine functions to describe the instantaneous current through and voltage across the load. Then apply the above to those functions to find average and RMS values.

11. Oct 6, 2013

### meBigGuy

How did you derive the equation for Vp? (yours only works for pure sine wave)

12. Oct 6, 2013

### Physixs

I think there are discrepancies between my book and the advice you are all giving me in that my book is a non calculus based physics book :(

We dont have any integrals or derivatives... just those equations to work with and somehow, with those equations, I have to answer this question. I havent had calc in years so I can only kind of follow along with what you are saying.

But it is very helpful to know that the formulas i am using are only for pure sine waves :)

13. Oct 6, 2013

### Drakkith

Staff Emeritus
What's the title of your book?

14. Oct 7, 2013

### meBigGuy

I'm only being logical here. Doesn't it seem the average and peak voltages would be linear sums of the components? The RMS is different in that it is about power/energy and is proportional to the square of the voltage (Power = (V^2)/R )

I would guess that Vp = Vp(sine wave) + Vdc (that is, the sum of those two components)

I would guess that Vavg = Vavg(sinewave) + Vavg(Vdc) (again, the sum of those two components.

I would say Vrms is as I said previously.

15. Oct 7, 2013

### Physixs

Thank you so much for all your help guys. It means a lot.

@drakkith
Title of book is Electric Circuit Fundamentals.

@Big Guy
When you say Vp(sinewave) do you mean the instantaneous voltage formula?
Also, it makes sense to me that the AC signal would cancel itself out (when the DC offset is high enough to offset the reverse alternations) Out of curiosity, why is that not the case (I know its not since the problems is asking for the avg voltage - which means the variations from 350V to 50V must mean something.

If I do, Vavg of 150 AC, I get about 95.5V (or can I not do this because the order in which I do this matters... in that I can't take the avg of the sine wave prior to its series combination with the DC - i think this is what you were saying before by saying it was no longer a "pure" sinewave

the avg of dc is just the dc... which would still be 200V? No?

(95.5^2 + 200^2 )^.5 = 221V.

221V/147 = 1.5A
1.5(100) = 150V - which is not the answer in the back of the book. :(

16. Oct 7, 2013

### sophiecentaur

You can work this out for yourself without integrating, explicitly, with calculus. A spreadsheet will give you the answer if you enter the values for the volts in small time steps (you could break the waveform down into, say 20 samples) over the AC cycle and then square and add them. Take the root of the answer and you will get the RMS value, to a close approximation. This is numerical integration and is a valid way of getting an answer.

17. Aug 12, 2015

### TENgineer

I realize this is an old post, but the questions were never answered, at least no correctly.
150Vpp on a 200V DC level through R1 = 470 ohms through RL = 100 ohms to ground.
Draw the total voltage across RL.
Determine the maximum current through RL.
And the average across RL. Does not state voltage or current, so I'll provide both.
First, 150VAC Peak-Peak is 75V Peak. Since the AC signal is on a DC level, the maximum voltage during the positive half is 275 V peak and 125 V peak during the negative half.
Determine the maximum current through RL: I = E/R = 275/(R1 + RL) = 275/570 = 0.482456 Amps or 482.456 mA.
The current through RL is the same as the current through R1.
Minimum current (will need this later): 125/570 = 0.21993 or 219.93 mA
The average across RL.
Average voltage of the circuit: 200V = (275 + 125)/2 or the DC level.
Average Current: I = E/R = 200/570 = 0.350877 A or 350.877 mA.
Average Voltage: E = I x R = 0.350877 x 100 = 35.088 V
Draw the total voltage across RL.
Max. E = I x R = 0.482456 x 100 = 48.246 V
Min. 0.350877 x 100 = 21.93 V
Average (DC level) = (48.246 + 21.93)/2 = 70.176/2 = 35.088V
The waveform across RL will be an AC signal that is 26.316 Vpp (13.158 V peak) on a 35.088 VDC level.