# Voltage doubled on capacitor

I try to explain what I see when I unplug capacitor from below circuit, and put it back inversely. So after the capacitor has been flipped voltmeter shows +6V and slowly dischares to 1,5V and stops there. It may seems to be normal at first but when I strat to think what happened I cannot understand it. Let's go back to the beginning, it is what the circuit below depicts. Capacitor is charged to 3V thru 10Mohms of voltmetter, so it has equal amaount of charge on each plate but with diffrenet signs. So one plate has X columbs of positive charge and other X columbs of negative charge. Now after we unplug capacitor and put it back inversly we have a situaltion in which negatively charged plate is now connected to positive terminal of battery and positive one to negative terminal of battery. So what happens? As I said, multimeter shows 6V and slowly discharges to 1.5V. First why 6V? If the negative cap plate is connected to positive battery terminal it should dissolve all negative cherges it has previously and gain these positive ones making this closer to the other plate charge therefore lowering the voltage. Why it isn't the case? Why then it stops discharging at 1.5V?

• Are you describing a real circuit or a just a simulation? electronics.stackexchange.com/questions/338838/… and this electronics.stackexchange.com/questions/301921/…
– G36
Jun 1, 2018 at 18:31
• Try using KVL on the new circuit immediately after the capacitor is re-inserted, and you'll understand the 6 V part. Why it stops discharging at 1.5 V is probably down to some detail of how the simulator models capacitors or meters that won't be obvious without a lot of digging. Jun 1, 2018 at 18:32
• What I am describing is real circuit. Jun 1, 2018 at 19:31
• @DannyS Haven't you use a polarized capacitor in this experiment? Because you can get such strange results if you use for example an electrolytic capacitor.
– G36
Jun 2, 2018 at 11:13
• @G36, It looks like I damaged the cap by flipping it back and forth! I figured that out because it no longer charges to Vcc if connected right to it. Now the question is what caused it? Jun 2, 2018 at 18:22

You can consider the capacitor as a (very tiny) rechargeable battery.

In your circuit, the capacitor will charge to 3 volts through the resistance of the voltmeter. When you reverse the capacitor connections, you effectively have two, three volt batteries in series, so the voltmeter will initially read 6 Volts.

The capacitor will then discharge, then re-charge with the opposite polarity. I would expect the meter to eventually read Zero volts, but there may be some leakage in the capacitor, or some other effect, that causes the meter to settle at 1.5 volts.

• with a 10 Meg in series, could it simply be that the 5 tau is much larger than the actual simulation time? Jun 2, 2018 at 0:18
• @SimonMarcoux: no, I don't think it is due to the time contstant - the OP implies that the voltage settles at 1.5 volts. Jun 2, 2018 at 0:40
• I think that you are probably right on this one, but not knowing the simulation parameters...It is the only logical idea that popped into my mind at that point to explain the remaining 1.5 volts. Correction: apparently the readings are from a real circuit and not some kind of simulator. Disregard my previous comment! Jun 2, 2018 at 0:43
• Well I wasn't clear enough - this problem is not about simulation. All I described happens in REAL circuit. Jun 2, 2018 at 8:21
• Why it stops at 1.5V? Does it mean that constantly current flow thru it? Jun 2, 2018 at 14:52

Once the capacitor is charged, it's a 3V source with a polarity opposite the battery, so the net voltage is 0. If you flip it around, it's another 3V source in series with the battery and with the same polarity, so they add to 6V.

That said, why it only discharges to 1.5V is more of a mystery. Is the cap polarized? An electrolytic/tantalum might leak in the reverse direction, but one doesn't often encounter these types in the sub-uF range.