Three condensers each of capacitance 2F are put in series. The resultant capacitance is

Options

A. 6 F
B. `(3)/(2)F`
C. `(2)/(3)F`
D. 5 F

Answer

To find the resultant capacitance of three capacitors connected in series, we can follow these steps:

### Step 1: Understand the formula for capacitors in series
When capacitors are connected in series, the reciprocal of the total capacitance (C_eq) is equal to the sum of the reciprocals of the individual capacitances. The formula is given by:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

### Step 2: Identify the values of the capacitors
In this case, we have three capacitors, each with a capacitance of 2 F. Therefore, we can denote:

\[ C_1 = C_2 = C_3 = 2 \, \text{F} \]

### Step 3: Substitute the values into the formula
Substituting the values into the formula, we get:

\[ \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \]

### Step 4: Simplify the equation
Adding the fractions gives:

\[ \frac{1}{C_{eq}} = \frac{3}{2} \]

### Step 5: Solve for C_eq
To find C_eq, we take the reciprocal of both sides:

\[ C_{eq} = \frac{2}{3} \, \text{F} \]

### Final Answer
Thus, the resultant capacitance of the three capacitors in series is:

\[ C_{eq} = \frac{2}{3} \, \text{F} \]

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