Effective capacitance of parallel combination of two capacitors C1 and C2 is 10 uF. When these capac

Q: Why do we use $U = \frac{1}{2}CV^2$ instead of $U = \frac{Q^2}{2C}$?

Because the problem specifies that the capacitors are connected to a constant voltage source ($1\text{ V}$), making the $V$ term constant and easier to compare. If they were connected in series, the charge $Q$ would be constant, and the other formula would be more useful.

Q: Why do capacitances add up in parallel?

In parallel, the effective plate area increases. Since capacitance is directly proportional to the plate area ($C = \frac{\epsilon_0 A}{d}$), connecting them in parallel directly increases the capacity to store charge for a given voltage.

Q: What if the energy relation was given for a series connection?

In a series connection, the charge $Q$ is constant across all capacitors. In that case, we would use $U = \frac{Q^2}{2C}$ to compare their energies, which would mean energy is inversely proportional to capacitance.

Q: Can the effective series capacitance ever be larger than the individual capacitors?

No, the equivalent capacitance in series is always strictly less than the smallest individual capacitor in the combination. This is a fundamental property of adding reciprocals.

Q: Is it possible for $C_1$ to be $8 \mu\text{F}$ and $C_2$ to be $2 \mu\text{F}$?

Based on the specific condition $U_2 = 4U_1$, $C_2$ must be $4$ times $C_1$. So $C_2$ must be the larger one. However, for the final series calculation, the order doesn't matter because multiplication and addition are commutative!

Question

Effective capacitance of parallel combination of two capacitors $C_1$ and $C_2$ is $10 \mu	ext{F}$. When these capacitors are individually connected to a voltage source of $1	ext{ V}$, the energy stored in the capacitor $C_2$ is 4 times that of $C_1$. If these capacitors are connected in series, then their effective capacitance will be

Options:

(A) $4.2 \mu	ext{F}$

(B) $3.2 \mu	ext{F}$

(C) $1.6 \mu	ext{F}$

(D) $8.4 \mu	ext{F}$

Sigma Percentile · Accepted Answer

Welcome to an exciting journey into the world of electrostatics! Today, we are going to unravel a classic problem involving the combination of capacitors. This problem is a beautiful blend of basic circuit theory and energy concepts. Let's dive right in and break it down step by step.

### Analyzing the Setup

Imagine you have two water tanks, which we will call our capacitors, $C_1$ and $C_2$. The problem states that when these two capacitors are connected in parallel, their effective capacitance is $10 \mu	ext{F}$.

What does a parallel connection mean physically? When capacitors are connected in parallel, it is akin to placing two tanks side by side and connecting their bases. The total capacity to hold water (or charge) simply adds up. Mathematically, this is expressed as:

$$C_p = C_1 + C_2 = 10 \mu	ext{F}$$

This is our first crucial piece of information. But we have two unknowns, $C_1$ and $C_2$, and only one equation. We need another clue to solve this puzzle.

### The Master Equation

The problem provides a fascinating second clue: when these capacitors are individually connected to a $1	ext{ V}$ voltage source, the energy stored in $C_2$ is exactly $4$ times the energy stored in $C_1$.

To use this clue, we need to recall the formula for the energy stored in a capacitor. The energy $U$ is given by:

$$U = \frac{1}{2}CV^2$$

Since both capacitors are connected to the same $1	ext{ V}$ source, the voltage $V$ is constant for both. Let's set up the equation based on the given condition:

$$U_2 = 4U_1$$

Substituting our energy formula into this relation, we get:

$$\frac{1}{2}C_2 V^2 = 4 	imes \frac{1}{2}C_1 V^2$$

### Unlocking the Capacitance Values

Now comes the elegant part. Because the voltage $V$ is the same for both capacitors, the $V^2$ terms and the $\frac{1}{2}$ terms cancel out perfectly on both sides of the equation. We are left with a beautifully simple relationship:

$$C_2 = 4C_1$$

This tells us that capacitor $C_2$ is four times larger than $C_1$. Now, we can substitute this relationship back into our initial parallel combination equation:

$$C_1 + 4C_1 = 10 \mu	ext{F}$$

$$5C_1 = 10 \mu	ext{F}$$

Solving for $C_1$, we find:

$$C_1 = 2 \mu	ext{F}$$

And since $C_2$ is four times $C_1$, we easily determine:

$$C_2 = 8 \mu	ext{F}$$

### Final Calculation

We have successfully found the individual values of our capacitors! The final step is to determine their effective capacitance when they are connected in **series**.

When capacitors are in series, the effective capacitance $C_s$ is found using the reciprocal formula, which for two capacitors simplifies to the "product over sum" rule:

$$C_s = \frac{C_1 C_2}{C_1 + C_2}$$

Let's plug in our values:

$$C_s = \frac{2 	imes 8}{2 + 8}$$

$$C_s = \frac{16}{10}$$

$$C_s = 1.6 \mu	ext{F}$$

And there we have it! The effective capacitance when connected in series is $1.6 \mu	ext{F}$.

Notice a universal rule here: the equivalent capacitance in a series circuit is always strictly less than the smallest individual capacitor in the combination (in this case, $1.6 \mu	ext{F}$ is less than $2 \mu	ext{F}$). This is a great way to quickly verify if your final answer makes physical sense!

Select Answer:

Visualized Solution

$C_{p} = C_{1} + C_{2}$

$U = \frac{1}{2} C V^{2}$

$U_{2} = 4 U_{1}$

$C_{1} and C_{2}$

$C_{s} = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$

\text{Conclusion}

The Sigma Insight: Combination of Capacitors

Analyzing the Setup

The Master Equation

Unlocking the Capacitance Values

Final Calculation

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