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I have the following circuit in LTSPice. I can see +- 3.5 uA of current at various nodes. I should be able to calculate the 3.5 uA given the R and the C. I note that the sim output is sinusoidal. If there were no battery in the circuit, I would not be posting this question.

So what is the question....

My question is, in a simple easy to understand chunk of words, what is the effect of the battery on the circuit and why?

I understand that the capacitor will pass ac current. I understand the capacitor will block dc. I can look at the RC and frequency and can (probably) figure out how much charge the capacitor accumulates between cycles.

The basic foundation I am stuck on is the following....

At the peak of the first positive cycle, both sides of the capacitor should have 10 volts on it and hence no current or charge occurring. Battey always supplies 10 volts DC. Prior to this point, maybe it discharges? If so, where, into the ac source?

When the AC flips polarity, it looks to me like a signal rising to 10 volts is going into a battery, which looks to be in series, so 20 volts on the capacitor this time?

In ltspice, i look at the voltage between v1 and r3, it is at a constant 10 volts. I would think it should show a rise to volts?

In ltspice, i look at the voltage between r3 and c1 and it appeats to have a sinusoidal signal on it on 10 volts varion +- 15 millivolts.

Why is this sinusoidal and not showing more current (20 volts) on one side and less current (10 volts) on the other?

Thanks for some basic clarification.

enter image description here

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  • \$\begingroup\$ more current (20 volts) on one side and less current (10 volts) on the other? - voltage is not current. 100pf at 600Hz represents an impedance of about 2.65 Megohms corresponding to about 3.7uA at 10V. " i look at the voltage between r3 and c3 " There is no c3 in your circuit. \$\endgroup\$
    – JIm Dearden
    Commented Mar 5, 2017 at 15:50
  • \$\begingroup\$ I meant that on the cycle of the circuit where 10 volts is present at a maximum, on the capacitor, less charge, and current should show on the capacitor, on the cycle of the circuit that has 20 volts on the capacitor, because the ac voltage is in series with the battery, more charge, and current should show on the capacitor. When I look in ltspice, the 3.7ua is perfectly symmetrical. I meant c1, not c3. I will change that. Between c1 and r3, there is fluxation about the 10 volt level. Just wondering why not between V1 and r3 also? Among other not intuitave aspects... like a missing 20v? \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Mar 5, 2017 at 16:16
  • \$\begingroup\$ So what is V1 doing and why? If I get rid of it, I see that c1, when connected to ground shows an alternating current through it, but no voltage on the wire from c1 and ground. I would think it should alternate +- 10v. When v1 is connected, as shown, there is a signal about 10v. is v1 or v2 charging the cap and why/where? if v1 charges it, then it must only charge when v2 is not at its peak of 10v, does it discharge? \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Mar 6, 2017 at 21:33
  • \$\begingroup\$ @JImDearden Do you have just a few comments on this question? I now understand that by definition, between V1 and R3 will be a constant 10V. Because v2 is fluctuating between +- 10v, I would think that because of v1 being constant, v2 sees a 0-20v oscillation at 600hz? How is v1 really affecting this? Thanks \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Mar 11, 2017 at 4:20

2 Answers 2

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V1 is a constant-voltage source, so, by definition, the voltage between V1 and R3 will be 10 volts, regardless of anything else in the circuit.

The voltage between C1 and R3 will vary around 10 volts, as C1 will be charged to 10 volts by V1, and the AC voltage from V2 will be coupled to the C1/R3 point by C1. If you use a larger capacitor (perhaps 0.1 - 1 uF) you should see the voltage between C1 and R3 varying between 0 and +20 volts.

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  • \$\begingroup\$ Why isn't C1 charged a little by V2? I understand that V2 alternates. So, I assume you are saying that 100% of the charge/discharge of C1 by V2 gets passed to C1/R3. Is this 100% whats referred to as coupling? Could 90% be coupled and 10% charge C1 by V2 if the size of C is changed? You say that C1 is charged by V1.. It must be only when V2 goes below 10V? If so, C1 can also discharge into the V2 source from a prior charge and V1 somehow couples into V2? Is all this so? \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Mar 6, 2017 at 14:47
  • \$\begingroup\$ Thanks , Can you just provide a tiny bit more detail with regard to my above comment? I guess I'm wondering more about what effect does v1 have on this circuit and why/how? \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Mar 10, 2017 at 21:07
  • \$\begingroup\$ If V2 is at zero volts, V1 will charge C1 to 10 volts through R3. If the C1/R3 time constant is sufficiently long, when V2 is a 10 V P-P sine wave, C1 will remain charged to 10 volts, and the junction between C1 and R3 will therefore swing between zero and 20 volts. If the C1/R3 time constant is less than a few cycles, C1 will partially charge and discharge each cycle, so the C1/R3 junction will swing less than +/-10 volts from +10 volts. \$\endgroup\$
    – Peter Bennett
    Commented Mar 10, 2017 at 21:33
  • \$\begingroup\$ Thanks @PeterBennett v2 hits 0 1200 times a second. It seems that v1 will always be able to charge c1 just a little, and, given enough time, c1 will become fully charged? When v2 goes to -10v, shouldn't it charge quicker from v1? If that is so, then the next thing that happens seems to depend on the time constant as you mention. I only see swings that reach 20v when i check the difference between the c1/r1 junction and the + lead on the v2 source, which alternates betwen +- 10v. Ground seems to always be at 0 in ltspice. \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Mar 13, 2017 at 14:30
  • \$\begingroup\$ I could have sworn I have seen ac circuits described where the + lead on V2 switches to 0 and the - lead switches to 10v and vice versa instead of having ground always remaining at at 0 and the + lead switching (on v2) . I suppose thats a differet way of thinking of it? But, in this circuit, ground is always at 0 and the + lead switches +- 10. Is that so? is that because v1 is connected to ground? Thats only where I see +- 20v swings ..... between the + on V2 and C1/R3 junction. \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Mar 13, 2017 at 14:31
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After a bit of research, the thing that helped me most in "closure" on this question is the superposition theorem. This is the theorem in electronics that allows one to analyze the contributions of each power source and add the separate layers to get the total picture/analysis of the crcuit. It works for linear circuits only. After learning this, it helped in perfectly answering this question (in my mind).

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