Why is the initial current in this capacitor not continuous?

Question

I am working on this circuit:

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?

Answer 1

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.

Answer 2

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.