Question

Three capacitors C\(_1\) = 2 \(\mu\)F, C\(_2\) = 6 \(\mu\)F and C\(_3\) = 12 \(\mu\)F are connected as shown in figure. Find the ratio of the charges on capacitors C\(_1\), C\(_2\) and C\(_3\) respectively: 

 

• A. 3 : 4 : 4
• B. 2 : 3 : 3
• C. 2 : 1 : 1
• D. 1 : 2 : 2

Hint:

When dealing with mixed series-parallel capacitor circuits, simplify the circuit step-by-step. First, identify the simplest series or parallel combinations and replace them with their equivalent capacitance. Then, work your way outwards until you have a single equivalent capacitor. To find charges/voltages, work backwards from the simplified circuit to the original.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a circuit with three capacitors and need to find the ratio of the charges stored on each capacitor. To do this, we must first analyze the circuit to see how the capacitors are connected (in series or parallel).
Step 2: Key Formula or Approach:
1. Capacitors in Series: When capacitors are in series, the charge on each capacitor is the same. The equivalent capacitance \( C_{\text{eq}} \) is given by \( \frac{1}{C_{\text{eq}}} = \frac{1}{C_a} + \frac{1}{C_b} + \dots \)
2. Capacitors in Parallel: When capacitors are in parallel, the voltage across each capacitor is the same. The equivalent capacitance is the sum \( C_{\text{eq}} = C_a + C_b + \dots \)
3. Charge on a Capacitor: The charge \(Q\) on a capacitor is given by \( Q = CV \), where \(V\) is the voltage across it.
Step 3: Detailed Explanation:
Circuit Analysis:
From the given figure, we can see that:
- Capacitors \( C_2 \) and \( C_3 \) are connected in series.
- This series combination of \( C_2 \) and \( C_3 \) is connected in parallel with capacitor \( C_1 \).
Calculate Equivalent Capacitance of the Series Combination:
Let's find the equivalent capacitance, \( C_{23} \), for \( C_2 \) and \( C_3 \).
\[ \frac{1}{C_{23}} = \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{6 \, \mu\text{F}} + \frac{1}{12 \, \mu\text{F}} \] \[ \frac{1}{C_{23}} = \frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4} \] \[ C_{23} = 4 \, \mu\text{F} \] Calculate Charges:
Let \(V\) be the total voltage applied across the entire circuit (between points A and B).
- The capacitor \( C_1 \) is in parallel with the combination \( C_{23} \). Therefore, the voltage across \( C_1 \) is \(V\), and the voltage across the combination \( C_{23} \) is also \(V\).
Charge on \( C_1 \) is \( Q_1 \):
\[ Q_1 = C_1 V = (2 \, \mu\text{F}) \times V = 2V \, \mu\text{C} \] - For the series combination \( C_2 \) and \( C_3 \), the total charge stored on the combination is \( Q_{23} \).
\[ Q_{23} = C_{23} V = (4 \, \mu\text{F}) \times V = 4V \, \mu\text{C} \] Since \( C_2 \) and \( C_3 \) are in series, the charge on each of them is the same and equal to the charge on their equivalent capacitor.
Therefore, \( Q_2 = Q_3 = Q_{23} \).
\[ Q_2 = 4V \, \mu\text{C} \] \[ Q_3 = 4V \, \mu\text{C} \] Find the Ratio:
We need to find the ratio \( Q_1 : Q_2 : Q_3 \).
\[ Q_1 : Q_2 : Q_3 = 2V : 4V : 4V \] We can cancel \(V\) from all terms:
\[ 2 : 4 : 4 \] Dividing all terms by the greatest common divisor, which is 2:
\[ 1 : 2 : 2 \] Step 4: Final Answer:
The ratio of the charges on capacitors \( C_1, C_2, \) and \( C_3 \) is 1 : 2 : 2.