Two capacitors are in parallel and when connected to a source of 3000 V, store 250 J of energy. When they are connected in series to the same source, the energy stored decreases by 190 J for the same potential. Their capacities are in the ratio

Options

A. `3: 2`
B. ` 2:7`
C. `4:3`
D. `3:5`

Answer

To solve the problem step by step, we need to analyze the energy stored in capacitors when they are connected in parallel and in series.

### Step 1: Energy Stored in Parallel
When two capacitors \( C_1 \) and \( C_2 \) are connected in parallel, the equivalent capacitance \( C_P \) is given by:
\[ C_P = C_1 + C_2 \]
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
Given that the energy stored when connected in parallel is 250 J and the voltage is 3000 V, we can write:
\[ 250 = \frac{1}{2} C_P (3000)^2 \]
This simplifies to:
\[ C_P = \frac{2 \times 250}{(3000)^2} = \frac{500}{9000000} = \frac{1}{18000} \text{ F} \]
Thus, we have:
\[ C_1 + C_2 = \frac{1}{18000} \text{ F} \quad \text{(Equation 1)} \]

### Step 2: Energy Stored in Series
When the same capacitors are connected in series, the equivalent capacitance \( C_S \) is given by:
\[ \frac{1}{C_S} = \frac{1}{C_1} + \frac{1}{C_2} \]
The energy stored in series is given by:
\[ E_S = \frac{1}{2} C_S V^2 \]
We know that the energy stored in series decreases by 190 J compared to the parallel connection:
\[ E_S = 250 - 190 = 60 \text{ J} \]
Thus, we can write:
\[ 60 = \frac{1}{2} C_S (3000)^2 \]
This simplifies to:
\[ C_S = \frac{2 \times 60}{(3000)^2} = \frac{120}{9000000} = \frac{1}{75000} \text{ F} \]
Thus, we have:
\[ C_S = \frac{C_1 C_2}{C_1 + C_2} = \frac{1}{75000} \quad \text{(Equation 2)} \]

### Step 3: Relating Equations
From Equation 1, we have:
\[ C_1 + C_2 = \frac{1}{18000} \]
From Equation 2, we can express \( C_1 C_2 \):
\[ C_S = \frac{C_1 C_2}{C_1 + C_2} \implies C_1 C_2 = C_S (C_1 + C_2) \]
Substituting the values:
\[ C_1 C_2 = \frac{1}{75000} \cdot \frac{1}{18000} \]
Calculating this gives:
\[ C_1 C_2 = \frac{1}{1350000000} \text{ F}^2 \]

### Step 4: Finding the Capacitor Ratio
Let \( C_1 = x \) and \( C_2 = y \). We have:
1. \( x + y = \frac{1}{18000} \)
2. \( xy = \frac{1}{1350000000} \)

Using the quadratic equation \( t^2 - (x+y)t + xy = 0 \):
\[ t^2 - \frac{1}{18000}t + \frac{1}{1350000000} = 0 \]
The roots of this equation will give us \( C_1 \) and \( C_2 \).

### Step 5: Solving for the Capacitor Ratio
Using the quadratic formula:
\[ t = \frac{\frac{1}{18000} \pm \sqrt{\left(\frac{1}{18000}\right)^2 - 4 \cdot \frac{1}{1350000000}}}{2} \]
Calculating the roots will yield \( C_1 \) and \( C_2 \), and we can find their ratio:
\[ \frac{C_1}{C_2} = \frac{3}{2} \]

### Final Answer
The ratio of the capacities \( C_1 : C_2 \) is \( 3 : 2 \).