Question

Three capacitors are connected to a battery as shown in figure. The ratio of charge on capacitors $C_3$ and $C_1$ is ______.

(Assuming standard bridge configuration where $C_1$ is in series with parallel $C_2, C_3$ based on OCR artifacts, or the visual Delta where battery connects to the base)

• A. 1.5
• B. 2.5
• C. 3.5
• D. 4.5

Hint:

In series, Charge ($Q$) is constant across all capacitors. In parallel, Voltage ($V$) is constant across all branches. Identifying these branches correctly is 90% of the battle in capacitor networks!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We have a circuit with three capacitors. Based on standard diagrams of this specific classic problem (a triangle Delta configuration where external lines connect to the bottom two vertices), the battery is connected directly across the base capacitor ($C_3$), placing it in parallel with the series combination of the top two capacitors ($C_1$ and $C_2$).

Step 2: Detailed Explanation:
Let's define the capacitance values given in the image text:
$C_1 = 1C$
$C_2 = 2C$
$C_3 = 3C$
1. Determine the voltage across the branches:
The battery is connected across the nodes of $C_3$. Let the battery voltage be $V$.
- The voltage across the bottom branch ($C_3$) is exactly $V$.
- The voltage across the top branch (the series combination of $C_1$ and $C_2$) is also exactly $V$.
2. Calculate the charge on $C_3$ ($Q_3$):
Using the basic capacitor formula $Q = CV$:
$Q_3 = C_3 \times V$
$Q_3 = 3C \times V = 3CV$ --- (Equation 1)
3. Calculate the charge on $C_1$ ($Q_1$):
Capacitors $C_1$ and $C_2$ are strictly in series. In a series circuit, the equivalent capacitance ($C_{eq}$) is:
$C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2}$
$C_{eq} = \frac{(1C) \times (2C)}{1C + 2C} = \frac{2C^2}{3C} = \frac{2}{3}C$
The total charge flowing into a series branch is the same as the charge on each individual capacitor within that branch. Therefore, $Q_1 = Q_2 = Q_{\text{series branch}}$.
$Q_1 = C_{eq} \times V$
$Q_1 = \left(\frac{2}{3}C\right) V = \frac{2}{3}CV$ --- (Equation 2)
4. Calculate the required ratio ($Q_3 / Q_1$):
$\text{Ratio} = \frac{Q_3}{Q_1} = \frac{3CV}{\frac{2}{3}CV}$
The $CV$ terms cancel out entirely:
$\text{Ratio} = \frac{3}{\frac{2}{3}} = 3 \times \frac{3}{2} = \frac{9}{2} = 4.5$

Step 3: Final Answer:
The ratio of charge is 4.5, corresponding to the corrected option (d).