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When I place a voltmeter across a capacitor, the reading comes out incorrect. Any ideas why? Am I doing something wrong, or is the simulator unable to calculate this correctly. I calculate 14.9V and 11.1V. The voltmeter reading is 13V and 13V. https://www.circuitlab.com/circuit/m4ht87/giancoli-circuit-37-38/

In CL, just as in real life;

i) voltmeters have finite resistances across their terminals;

ii) caps have very large but finite resistances in parallel with them.

See these threads:

https://www.circuitlab.com/forums/analog-design/topic/b95fc6f8/can-a-voltmeter-be-given-infinite-resistance-for-education/

https://www.circuitlab.com/forums/support/topic/4eqt8q32/voltage-calculations/

@Signality: Can’t stop re-reading your posting ! ;-)

But I can’t understand the OP’s concern, how to “calculate 14.9V and 11.1V” using a DC source. How come?

Run DC solver, gives 13V (as expected, but bear in mind this value may not be true in reality, depends on cap’s type, age, …).

With two voltmeters they read 13V each, one voltmeter gives 26V (= Signality's answer together with CL's ideal reality).

Regards, Sancho

"But I can’t understand the OP’s concern, how to “calculate 14.9V and 11.1V” using a DC source. How come?"

The underlying assumption is that the voltage across two ideal caps in series - each initially having a charge of zero Coulombs - which have a voltage of V applied across them in series, will be divide as a potential divider but in the inverse ratio of their capacitances. (Think of capacitive impedance rather than the actual capacitance.)

The smaller cap will charge to the higher voltage so if the caps are Ca = 1F and Ca =2F and the total voltage V = 1V then V_Ca = 1/3V and V_Cb = 2/3V.

That's where the quoted voltages come from:

V_C1 = 4/7*26V

V_C2 = 3/7*26V

Is good yes?

Uuups, good, indeed, you see, my brain isn’t ideally working, especially with similar caps … Seems it would need 4uF in series to 3nF to start thinking.

:-(

Regards, Sancho