CURRENT ChYU 3 - CHATUP707

Detailed Physics Solution – Maximum Potential

Solution: Maximum Potential Difference with Switched Capacitors

The objective is to maximize the potential difference across a terminal $AB$ using a battery of voltage $V_0$ and two capacitors $C_1$ and $C_2$. This is achieved in three distinct phases.

Phase 1: Recursive Charge Pumping (Charging $C_2$)

First, we use $C_1$ as a “charge pump.” In every cycle, $C_1$ is charged to $V_0$ by the battery, and then connected in series with the battery to push charge onto $C_2$.

Figure 1: The recursive loop. $C_1$ acts as a bucket, adding $V_0$ potential repeatedly.

Applying KVL to this loop in the steady state (where charge transfer stops):

\text{Loop Potential} = V_{\text{battery}} + V_{C1} – V_{C2} = 0 $$ $$ V_0 + V_0 – V_{C2} = 0 $$ $$ V_{C2} = 2V_0

Result 1: Capacitor $C_2$ is now charged to a potential of $2V_0$.

Phase 2: The Intermediate Step (Recharging $C_1$)

After $C_2$ is fully charged, $C_1$ is disconnected from the loop. It is then connected directly across the battery again to ensure it holds a fresh potential of $V_0$.
(This step provides the “third” voltage source for our final stack).

Figure 2: Refreshing $C_1$ to $V_0$ completely independent of $C_2$.

Result 2: Capacitor $C_1$ is now charged to a potential of $V_0$.

Phase 3: Final Stacking & Calculation

We now have three “batteries”: 1. The actual battery ($V_0$) 2. Capacitor $C_2$ ($2V_0$) 3. Capacitor $C_1$ ($V_0$)
We connect them all in loop, a small charge $q$ will flow to equalize the circuit.

Step 3a: Calculate Redistribution Charge ($q$)

We apply Kirchhoff’s Voltage Law (KVL) to the closed loop formed by the components:

V_0 + \frac{2C_2 V_0 – q}{C_2} – \frac{C_1 V_0 + q}{C_1} = 0 $$ Separating the fractions: $$ V_0 + 2V_0 – \frac{q}{C_2} – V_0 – \frac{q}{C_1} = 0 $$ $$ 2V_0 – q\left( \frac{1}{C_2} + \frac{1}{C_1} \right) = 0 $$ Solving for $q$: $$ q\left( \frac{C_1 + C_2}{C_1 C_2} \right) = 2V_0 \implies q = \frac{2V_0 C_1 C_2}{C_1 + C_2}

Step 3b: Calculate Total Potential ($V_{AB}$)

Figure 3: All components stacked in series.

The total potential across A and B is the sum of the potentials of the three components in series.

V_{AB} = V_{\text{batt}} + V_{C2}’ + V_{C1}’ $$ $$ V_{AB} = V_0 + \left( 2V_0 – \frac{q}{C_2} \right) + \left( V_0 + \frac{q}{C_1} \right) $$ $$ V_{AB} = 4V_0 + q\left( \frac{1}{C_1} – \frac{1}{C_2} \right)

Now we substitute the value of $q$ we found in Step 3a:

V_{AB} = 4V_0 + \left( \frac{2V_0 C_1 C_2}{C_1 + C_2} \right) \left( \frac{C_2 – C_1}{C_1 C_2} \right) $$ Canceling $C_1 C_2$: $$ V_{AB} = 4V_0 + \frac{2V_0 (C_2 – C_1)}{C_1 + C_2} $$ Finding a common denominator ($C_1 + C_2$): $$ V_{AB} = \frac{4V_0(C_1 + C_2) + 2V_0(C_2 – C_1)}{C_1 + C_2} $$ $$ V_{AB} = \frac{4V_0 C_1 + 4V_0 C_2 + 2V_0 C_2 – 2V_0 C_1}{C_1 + C_2} $$ $$ V_{AB} = \frac{2V_0 C_1 + 6V_0 C_2}{C_1 + C_2} $$ Factoring out $2V_0$: $$ V_{AB} = 2V_0 \left( \frac{C_1 + 3C_2}{C_1 + C_2} \right)

Substituting the given values: $V_0 = 1.5$ V, $C_1 = 2$ $\mu$F, $C_2 = 3$ $\mu$F:

$$ V_{max} = 2(1.5) \left( \frac{2 + 3(3)}{2 + 3} \right) $$ $$ V_{max} = 3 \left( \frac{11}{5} \right) = 3(2.2) $$ $$ \mathbf{V_{max} = 6.6 \text{ V}} $$