Capacitor In Parallel Interview Question | VLSI Design

Questions:

Two capacitors are connected in parallel through a switch. C1= 5uF, C2= 2uF. Initially switch is open,C1 is charged to 8 V. What happens if we close the switch? No loss in the wires and capacitors. 

Answer:

When two capacitors are connected in parallel through a switch, and initially, one of them is charged while the other is not, closing the switch will result in the redistribution of charge between the capacitors. Here is a step‐by‐step analysis of this particular scenario:

Initial Conditions ‐ Capacitor $C_1$ has a capacitance of $5\mu F$ and is initially charged to $8V$. ‐ Capacitor $C_2$ has a capacitance of $2\mu F$ and is initially uncharged.

Charge on Capacitors Initially, the charge $Q$ on capacitor $C_1$ can be calculated using the formula: $$Q_1=C_1\times V_1$$ $$Q_1=5\mu F\times 8V=40\mu C$$

Capacitor $C_2$ is uncharged, so: $$Q_2=0\mu C\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}$$

Closing the Switch When the switch is closed, the capacitors will share their charges until they reach the same voltage, as they are in parallel. The final voltage $V_f$ will be the same across both capacitors.

The total charge in the system before closing the switch is: $$Q_{total}=Q_1+Q_2=40\mu C+0\mu C=40\mu C\mathrm{<br>}$$

After closing the switch, this total charge $Q_{total}$ will be distributed between the two capacitors $C_1$ and $C_2$.

Final Voltage Calculation The total capacitance $C_{total}$ is the sum of the individual capacitances:

$$C_{total}=C_1+C_2=5\mu F+2\mu F=7\mu F$$

Using the conservation of charge: $$Q_{total}=C_{total}\times V_f$$

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Solving for $V_{\mathrm{<span\; style="font-size:\; 1.2em;">f</span>}}$$$=\frac{Q_{tota\mathrm{l\; /}}}{C_{total}}=\frac{40\mu \mathrm{C\; /}}{7\mu F}\approx 5.71V$$

Final Charges on Each Capacitor Now that we know the final voltage $V_f$ is approximately $5.71V$, we can find the final charge on each capacitor:

For $C_1$: $$Q_{1,f}=C_1\times V_f=5\mu F\times 5.71V\approx 28.57\mu C$$

For $C_2$: $$Q_{2,f}=C_2\times V_f=2\mu F\times 5.71V\approx 11.43\mu C$$

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Summary: When the switch is closed: ‐ The voltage across both capacitors will be approximately $5.71V$. ‐ The charge on $C_1$ will be approximately $28.57\mu C$. ‐ The charge on $C_2$ will be approximately $11.43\mu C$.

The system conserves charge, and the final charges on the capacitors reflect the redistribution based on their capacitances. The resulting voltage is a weighted average of the initial conditions, considering the different capacitance values.

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