# Closing a switch in series with a capacitor

Suppose we have the following circuit:

Such that for $t<0$ the switch M was open. If we close the switch at $t=0$ what will the voltage on the capacitor, $V_C$, be at $t=0^+$? What about $\dot V_C$ at $t=0^+$? Will there be a current passing through $R$ the moment the switch is closed?

I need to solve an ODE for a more complicated circuit which has this sub-circuit as a part of it, and I need initial conditions to solve for the ZIR case. I'm trying to figure out these initial conditions but I'm not sure. Here are my thoughts:

Before closing the switch, there will be a steady finite current $I$ in the circuit.The moment we close the switch, there can be no current passing through $R$, else there will be some finite voltage on the capacitor $\longrightarrow$ $\dot V_C$ will be infinite $\longrightarrow$ $I_C$=C$\cdot \dot V_C$ will be infinite which cannot be.

Therefore $V_C(t=0^+)=0$ and $\dot V_C(t=0^+)=I/C$. I'm not sure if what I said I correct, I mean we learned that if there is no impulse current (like Dirac's Delta function), the voltage on the capacitor will be continuous.Does this apply in the case too? I would really appreciate any help.

• The circuit, as drawn, cannot be solved conventionally. However, one can introduce a resistor $R_S$ in series with the voltage source (or capacitor) and then examine the limit of that solution as $R_S \rightarrow 0$. – Alfred Centauri Dec 27 '15 at 20:53
• My bad, there is actually a resistor in series with the Voltage Source. Does this mean what I said is true? Will the same analysis hold? Thanks – Dylan132 Dec 27 '15 at 21:29
• If you put a resistor $R_s$ in series with the voltage source then there will be a current flow proportional to $V/R_s$, which will then decrease exponentially because the capacitor is charging up, then the remaining current flow will be $V/(R+R_s)$ because of the two resistors in series. – CuriousOne Dec 27 '15 at 22:35

There are some mistakes with your assumptions. When $t<0$ a current $I =V/R$ will be flowing through resistor and no current would be flowing through capacitor. As soon we close the switch the capacitor will get charged instantaneously (yes it could lead to $I = \infty$ at $t=0$ but it can be avoided if even a small resistor is placed between capacitor and voltage source. And we can never get the resistance of those wires down to $0$). After $t=0$ a potential difference will develop across capacitor due to its charge. But a same current $I = V/R$ through the resistor as to satisfy Kirchhoff's loop rule in loop containing resistor and source. And because charge on capacitor will be constant after $t=0$ so will be the potential difference so $\dot{V_c}$ will be zero, not infinite and so will be the current through capacitor.

Let me tell you some notable points these should help you solve such problems from the competitive exam point of view:
=>A completely uncharged capacitor is like a wire, it offers 0 resistance
=>A fully charged capacitor is infinite resistance and no current flows through it
=>A battery that you have represented is an ideal battery and ideal batteries do not exist!!! To solve this question what you need to assume is that there is internal resistance of the cell that will be present.
=> The charging and discharging of capacitors is an exponential function which is very important from the physics point of view.

$V_{t} = V_{max} e^{\frac{-t}{RC}}$