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Actually i was practicing circuit analysis with Differential Equations. This question has some confusing parts like-

  1. In Steady State, t<0 switch is still open the 10mF capacitor is Charged to 5V and 0A current flows through it, now if a 8mF Uncharged capacitor(0V) is connected at time t=0 as show in picture then What will be Node voltage v(t) or voltage across both capacitors as they both are in parallel is it 0v or 5v? Because if the voltage at node A for t>0 is at 5v that means 8mF Cap. charged instantaneously and if the node voltage becomes same as the uncharged cap i.e. 0v then also 10mF cap goes from 5 to 0. someone please clarify

  2. There are two Initial Conditions, for 10mF v(0⁻) = 5v and for 8mF v(0⁻) = 0v, and the Variable is v(t), so value should i put in the end ? and i m getting different voltage expression for each initial condition. here's my work

$$\Rightarrow \dfrac{v\left( t\right) }{8}+10m\cdot \dfrac{dv\left( t\right) }{dt}+8m\cdot\dfrac{dv\left( t\right) }{dt}-\dfrac{5}{8}=0 ,\hspace{1cm} By \space KCL$$ $$\Rightarrow 18m\dfrac{dv\left( t\right) }{dt}+\dfrac{v}{8}=\dfrac{5}{8}$$ $$v\left( t\right) =A⋅e^{\dfrac{-t}{144⋅10^{-3}}}+5$$

It would we be really helpful for me if someone will give me some hints or physical reasoning or teach me how to handle these types of circuit situation instead of Direct Answer.

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    \$\begingroup\$ (1) it's indeterminate due to infinite current flow \$\endgroup\$
    – Andy aka
    Feb 9 at 13:58
  • \$\begingroup\$ This circuit can not be solved analytically because it violates our definitions of how capacitors behave and the meaning of parallel circuits. Connecting two ideal capacitors in parallel when they have different initial voltages requires infinite current for zero time. \$\endgroup\$ Feb 9 at 14:22
  • \$\begingroup\$ @ElliotAlderson Sir, could you please explain the this line "requires infinite current for zero time", what does it means ? \$\endgroup\$
    – fpsshubham
    Feb 9 at 14:43
  • \$\begingroup\$ @Andyaka, sir could you please elaborate it little bit ? and also is it solvable, Means can we find expression for i(t) and v(t) ? \$\endgroup\$
    – fpsshubham
    Feb 9 at 14:47
  • \$\begingroup\$ The current through a capacitor is proportional to the derivative of the voltage across it. What happens to the voltages across two capacitors that initially have different voltages and then you connect them in parallel. What is \$dV/dt\$ in this case? \$\endgroup\$ Feb 9 at 15:32

1 Answer 1

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Putting infinite instanteneous current between caps aside, the question is probably about charge redistributing between the two capacitors. So the question is really like "here is 5V cap of 10mF, how does the charge redistribute if we attach another 8mF cap to it.

We want to find total charge on capacitor C1, which will be the same as total charge on capacitors C1 and C2 once C2 is connected.

The formula for it says Q = C x V.

Two capacitors in parallel is just one larger capacitor. Which will have the same total charge as before.

C1 x V1 = Q = (C1 + C2) x V2

I hope it's obvious why voltage on two caps is the same when they're connected in parallel.

This charge distribution is valid strictly for time moment t=0, at t>0 the capacitors will be gaining extra charge via the resistor, where you have a typical RC-circuit, simply with starting condition that the capacitor is not at 0V. It will charge from V2 to 5V eventually. But I don't see explicit question about t>0, only about t=0.

Edit: perfect components, perfect connections assumed. "Spherical capacitors in a vacuum"

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  • \$\begingroup\$ ok so for t>0 how should i proceed? find differential equation by Kcl then solve it and the what condition should i apply ? please help \$\endgroup\$
    – fpsshubham
    Feb 10 at 9:14
  • \$\begingroup\$ @fpsshubham it's simple RC circuit with IC formulae you can find anywhere. Remember that voltage is relative. Charging from 0V to 5V is IDENTICAL to the last letter and digit to charging from 995V to 1000V and from -34V to -29V. In this case, when at t=0 voltage on caps immediately drops, you will have a classic RC circuit with capacitor(s) charging from V2 to 5V. \$\endgroup\$
    – Ilya
    Feb 10 at 9:16
  • \$\begingroup\$ Here you can find detailed explanation as well as the formula: electronics-tutorials.ws/rc/rc_1.html Remember to adjust for charging not from 0 to 5V, but from V2 to 5V. \$\endgroup\$
    – Ilya
    Feb 10 at 9:18
  • \$\begingroup\$ ok i'll give it a try! \$\endgroup\$
    – fpsshubham
    Feb 10 at 9:21
  • \$\begingroup\$ Let me know what your V2 is and how you apply the RC formula. I did some napkin math. \$\endgroup\$
    – Ilya
    Feb 10 at 9:22

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