Question:
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
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.
Updated On: Jun 16, 2026
- $4V_0/R_L$
- $V_0/R_0$
- $3V_0/R_0$
- $V_0/R_L$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The circuit shown is a classic Marx generator configuration.
Initially, the capacitors are charged in parallel to a voltage $V_0$. When all the switches are closed, the capacitors are reconfigured in series to discharge through a load resistor.
Step 2: Key Formula or Approach:
- When $n$ capacitors, each charged to voltage $V_0$, are connected in series, their voltages add up.
- The total equivalent voltage of the series combination is:
\[ V_{\text{total}} = n V_0 \]
- The current $I$ flowing through the load resistor $R_L$ at the instant of connection is given by Ohm's Law:
\[ I = \frac{V_{\text{total}}}{R_L} \]
Step 3: Detailed Explanation:
• The given circuit has 4 capacitors, each initially fully charged to the source voltage $V_0$ when the switches are open.
• When the switches $S$ are simultaneously closed, they connect the positive plate of one capacitor to the negative plate of the next.
• This places all 4 capacitors in a series configuration across the load resistor $R_L$.
• The total voltage across the load resistor at the instant of closing is:
\[ V_{\text{instant}} = 4V_0 \]
• Thus, the initial current flowing through the load resistor $R_L$ is:
\[ I = \frac{4V_0}{R_L} \]
Step 4: Final Answer:
The current flowing through the circuit at that instant is $4V_0/R_L$.
The circuit shown is a classic Marx generator configuration.
Initially, the capacitors are charged in parallel to a voltage $V_0$. When all the switches are closed, the capacitors are reconfigured in series to discharge through a load resistor.
Step 2: Key Formula or Approach:
- When $n$ capacitors, each charged to voltage $V_0$, are connected in series, their voltages add up.
- The total equivalent voltage of the series combination is:
\[ V_{\text{total}} = n V_0 \]
- The current $I$ flowing through the load resistor $R_L$ at the instant of connection is given by Ohm's Law:
\[ I = \frac{V_{\text{total}}}{R_L} \]
Step 3: Detailed Explanation:
• The given circuit has 4 capacitors, each initially fully charged to the source voltage $V_0$ when the switches are open.
• When the switches $S$ are simultaneously closed, they connect the positive plate of one capacitor to the negative plate of the next.
• This places all 4 capacitors in a series configuration across the load resistor $R_L$.
• The total voltage across the load resistor at the instant of closing is:
\[ V_{\text{instant}} = 4V_0 \]
• Thus, the initial current flowing through the load resistor $R_L$ is:
\[ I = \frac{4V_0}{R_L} \]
Step 4: Final Answer:
The current flowing through the circuit at that instant is $4V_0/R_L$.
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