All the capacitors in a given circuit are initially fully charged with all switches open. At a later time $t$, all the switches are simultaneously closed. The current flowing through the circuit at that instant is given by
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A Marx generator is a clever way to generate high-voltage pulses from a low-voltage DC supply by charging capacitors in parallel and discharging them in series.
Step 1: Understand the trick of the circuit. This is a Marx-type setup. The capacitors are charged side by side in parallel, each reaching the same voltage $V_0$.
Step 2: See what closing the switches does. When all switches close together, the wiring flips so the charged capacitors are suddenly stacked in series instead of parallel.
Step 3: Add the voltages in series. Stacking $n$ capacitors, each at $V_0$, in series makes their voltages add up to $n V_0$ across the chain at that first instant.
Step 4: Count the capacitors. From the figure there are four capacitors, so the combined driving voltage just after closing is $4 V_0$.
Step 5: Find the current through the load. At that instant the capacitors act like a battery of $4V_0$ pushing current through the load resistor $R_L$, so by Ohm's law \[ I = \frac{4 V_0}{R_L}. \]
Step 6: State the result. The momentary surge current is therefore four times what a single capacitor would drive. \[ \boxed{I = \dfrac{4 V_0}{R_L}} \]
