# Basic Capacitor Question To help Cement My Foundation Properly (LTSpice)

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. • 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. – JIm Dearden Mar 5 '17 at 15:50
• 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? – Jeffrey Edward Messikian Mar 5 '17 at 16:16
• 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? – Jeffrey Edward Messikian Mar 6 '17 at 21:33
• @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 – Jeffrey Edward Messikian Mar 11 '17 at 4:20