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I am working on this circuit:

enter image description here

Basically, \$V_1\$ is on all the time, and \$V_2\$ turns on at \$t=0\$. For the conditions at \$t=0^-\$ (the time just before zero), I can see why the voltage across the capacitor is equal to \$V_1\$. The current through the resistor at \$t=0^-\$ is zero, as no current would flow.

After the second voltage source turns on, my sources tell me that the current flowing through \$C\$ is equal to \$V_2 / R_2 \$. That is, the current instantaneously goes from 0 to \$V_2 / R_2 \$ at \$t=0\$.

What is going on here?

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    \$\begingroup\$ I can see why the voltage across the resistor is equal to V1 Are you talking about R2? If so, then I don't agree with this. You can't have zero current and have a finite voltage. That doesn't make sense. \$\endgroup\$ – KingDuken Oct 1 at 3:59
  • \$\begingroup\$ @KingDuken Thanks, I misspoke. I meant to say capacitor. \$\endgroup\$ – axsvl77 Oct 1 at 11:16
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At \$t=0^{-}\$ the current in the capacitor is zero, since it's an open circuit when only DC voltages are applied in the circuit.

At \$t=0\$, \$V_2\$ comes in, and so does the current through the capacitor, which is now \$\frac{V_2 - V_1}{R_2}\$ since at \$t=0\$ no current flows through the inductor branch.

Finally, remember that the capacitor current can change instantaneously, it's its voltage that cannot.

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  • \$\begingroup\$ Agreed, I think the key bit of information you mention is that at the instant V2 is connected the current through the inductor must still be zero, so whatever current flows through R2 is also the current through C. \$\endgroup\$ – Elliot Alderson Oct 1 at 15:52
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the current instantaneously goes from 0 to 𝑉2/𝑅2 at 𝑡=0.

There's nothing wrong with that. Current in a capacitor is completely arbitrary. It's the voltage that must be continuous.

Similarly, in an inductor, the voltage can be arbitrary, but the current must be continuous.

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