The equivalent capacitance of three capacitors of capacitance `C_(1):C_(2)` and `C_(3)` are connected in parallel is `12` units and product `C_(1).C_(2).C_(3) = 48`. When the capacitors `C_(1)` and `C_(2)` are connected in parallel, the equivalent capacitance is `6` units. then the capacitance are

Options

A. `2, 3, 7`
B. `1.5, 2.5, 8`
C. `1, 5, 6`
D. `4, 2, 6`

Answer

To solve the problem, we will follow a systematic approach using the given information about the capacitors.

### Step 1: Set up the equations based on the problem statement
We are given three capacitors with capacitances \( C_1, C_2, \) and \( C_3 \). The following equations can be derived from the problem:

1. The total capacitance when all three capacitors are connected in parallel:
\[ C_1 + C_2 + C_3 = 12 \quad \text{(Equation 1)} \]

2. The product of the capacitances:
\[ C_1 \cdot C_2 \cdot C_3 = 48 \quad \text{(Equation 2)} \]

3. The total capacitance when \( C_1 \) and \( C_2 \) are connected in parallel:
\[ C_1 + C_2 = 6 \quad \text{(Equation 3)} \]

### Step 2: Express \( C_3 \) in terms of \( C_1 \) and \( C_2 \)
From Equation 3, we can express \( C_3 \):
\[ C_3 = 12 - (C_1 + C_2) = 12 - 6 = 6 \quad \text{(Equation 4)} \]

### Step 3: Substitute \( C_3 \) into Equation 2
Now, substituting \( C_3 \) from Equation 4 into Equation 2:
\[ C_1 \cdot C_2 \cdot 6 = 48 \]
Dividing both sides by 6 gives:
\[ C_1 \cdot C_2 = 8 \quad \text{(Equation 5)} \]

### Step 4: Express \( C_1 \) in terms of \( C_2 \)
From Equation 5, we can express \( C_1 \):
\[ C_1 = \frac{8}{C_2} \quad \text{(Equation 6)} \]

### Step 5: Substitute \( C_1 \) into Equation 3
Substituting Equation 6 into Equation 3:
\[ \frac{8}{C_2} + C_2 = 6 \]
Multiplying through by \( C_2 \) to eliminate the fraction:
\[ 8 + C_2^2 = 6C_2 \]
Rearranging gives us a quadratic equation:
\[ C_2^2 - 6C_2 + 8 = 0 \quad \text{(Equation 7)} \]

### Step 6: Solve the quadratic equation
We can solve Equation 7 using the quadratic formula:
\[ C_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -6, c = 8 \):
\[ C_2 = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \]
\[ C_2 = \frac{6 \pm \sqrt{36 - 32}}{2} \]
\[ C_2 = \frac{6 \pm \sqrt{4}}{2} \]
\[ C_2 = \frac{6 \pm 2}{2} \]
This gives us two solutions:
\[ C_2 = \frac{8}{2} = 4 \quad \text{or} \quad C_2 = \frac{4}{2} = 2 \]

### Step 7: Find corresponding values of \( C_1 \)
Using Equation 6 to find \( C_1 \):
1. If \( C_2 = 4 \):
\[ C_1 = \frac{8}{4} = 2 \]
2. If \( C_2 = 2 \):
\[ C_1 = \frac{8}{2} = 4 \]

### Step 8: Find \( C_3 \)
From Equation 4, we already found:
\[ C_3 = 6 \]

### Final Result
Thus, the capacitances are:
- \( C_1 = 2, C_2 = 4, C_3 = 6 \) or
- \( C_1 = 4, C_2 = 2, C_3 = 6 \)

### Summary
The capacitances are \( C_1 = 2 \) units, \( C_2 = 4 \) units, and \( C_3 = 6 \) units.

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