In the given circuit, the charge on 4 μF capacitor will be
(a) 5.4 αC
(b) 9.6 αC
(c) 13.4 αC
(d) 24

Q: Why do we ignore the $3\mu\text{F}$ capacitor?

The $3\mu\text{F}$ capacitor is in a separate parallel branch connected directly to the battery. It draws its own charge independently and does not affect the potential difference or the charge distribution in the top branch.

Q: Why do we add the capacitances for the $1\mu\text{F}$ and $5\mu\text{F}$ capacitors?

Capacitors in parallel share the same potential difference. Their equivalent capacitance is simply the sum of their individual capacitances, $C_p = C_1 + C_2$.

Q: Why is the charge on the $4\mu\text{F}$ capacitor equal to the total charge of the upper branch?

In a series circuit, the charge on each component is identical. Since the $4\mu\text{F}$ capacitor is in series with the rest of the upper branch, the total charge flowing into that branch must accumulate on its plates.

Q: How would I find the charge on the $1\mu\text{F}$ capacitor?

First, find the voltage across the parallel combination by subtracting the voltage drop across the $4\mu\text{F}$ capacitor from the total $10\text{V}$. Then, use $Q = CV$ specifically for the $1\mu\text{F}$ capacitor with that new voltage.

Q: What if the battery had internal resistance?

If the battery had internal resistance, there would be a voltage drop across it when current flows. However, in a steady-state DC circuit with capacitors, no current flows after they are fully charged, so the internal resistance wouldn't affect the final voltages.

Question

In the given circuit, the charge on $4 \mu	ext{F}$ capacitor will be

Options:

(A) $5.4 \mu	ext{C}$

(B) $9.6 \mu	ext{C}$

(C) $13.4 \mu	ext{C}$

(D) $24 \mu	ext{C}$

Sigma Percentile · Accepted Answer

### Demystifying Capacitor Circuits: Finding the Charge on a Specific Capacitor

When faced with a complex-looking circuit, the best strategy is always to break it down into smaller, manageable pieces. In this problem, we are asked to find the exact charge stored on the $4\mu	ext{F}$ capacitor located in the top branch of the circuit.

Let's carefully observe the setup. We have a $10	ext{V}$ battery connected across two main parallel branches. The middle branch contains a single $3\mu	ext{F}$ capacitor, while the top branch is a combination of three different capacitors. Our focus will be entirely on this top branch.

### Analyzing the Setup

Notice the right side of the top branch. The $1\mu	ext{F}$ and $5\mu	ext{F}$ capacitors are connected in parallel. We know this because their plates are connected to the same two nodes, meaning they share the exact same potential difference.

For capacitors in parallel, we simply add their capacitances to find the equivalent capacitance.

$$C_p = C_1 + C_2$$

So, $1\mu	ext{F} + 5\mu	ext{F}$ gives us an equivalent capacitance of $6\mu	ext{F}$ for this section.

### The Master Equation

Now, look at the entire top branch as a whole. The $4\mu	ext{F}$ capacitor is in series with our newly found $6\mu	ext{F}$ equivalent capacitor. In a series connection, the equivalent capacitance is calculated using the "product over sum" rule.

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

Let's calculate this for our upper branch:

$$C_{upper} = \frac{4 	imes 6}{4 + 6} = \frac{24}{10} = 2.4\mu	ext{F}$$

This $2.4\mu	ext{F}$ represents the total equivalent capacitance of the entire upper branch.

### Final Calculation

Here is a crucial observation: The entire upper branch is connected directly in parallel with the $10	ext{V}$ battery. Therefore, the potential difference across the entire upper branch is exactly $10	ext{V}$. The middle $3\mu	ext{F}$ capacitor doesn't affect the voltage across this branch at all.

Now we can find the total charge supplied to the upper branch using the fundamental capacitor equation:

$$Q_{upper} = C_{upper} 	imes V_{upper}$$

$$Q_{upper} = 2.4\mu	ext{F} 	imes 10	ext{V} = 24\mu	ext{C}$$

Finally, remember a fundamental rule of series circuits: the charge on each capacitor in series is the same as the total charge of that branch. Since the $4\mu	ext{F}$ capacitor is in series with the rest of the branch, it must hold the entire $24\mu	ext{C}$ of charge.

$$Q_{4\mu	ext{F}} = 24\mu	ext{C}$$

**The correct answer is $24\mu	ext{C}$.**

Animated Solution for Physics - Electrostatics: In the given circuit, the charge on $4 μ F$ capacitor will be

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Visualized Solution

Circuit Analysis

Parallel Capacitors

Equivalent Parallel Capacitance

Series Capacitors

Equivalent Capacitance of Upper Branch

Potential Difference across Upper Branch

Charge on Upper Branch

Charge on $4 μ F$ Capacitor

Further Exploration

The Sigma Insight: Combination of Capacitors

Demystifying Capacitor Circuits

Analyzing the Setup

The Master Equation

Final Calculation

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Animated Solution for Physics - Electrostatics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the given circuit, the charge on 4μF capacitor will be