Two capacitors of capacitance C are connected in series. If one of them is filled with dielectric substance K, what is the effective capacitance?

Options

A. `(KC)/((1 + K))`
B. `C(K + 1)`
C. `(2KC)/((1 + K))`
D. None of these

Answer

To find the effective capacitance of two capacitors connected in series, where one of them is filled with a dielectric substance with dielectric constant \( K \), we can follow these steps:

### Step 1: Identify the capacitances
Let the capacitance of the first capacitor (without dielectric) be \( C_1 = C \). The second capacitor (with dielectric) will have its capacitance increased due to the dielectric. The capacitance of the second capacitor can be expressed as:
\[ C_2 = K \cdot C \]
where \( K \) is the dielectric constant.

### Step 2: Use the formula for capacitors in series
For capacitors in series, the effective capacitance \( C_{eff} \) can be calculated using the formula:
\[ \frac{1}{C_{eff}} = \frac{1}{C_1} + \frac{1}{C_2} \]

### Step 3: Substitute the values of \( C_1 \) and \( C_2 \)
Substituting the values we identified in Step 1 into the formula:
\[ \frac{1}{C_{eff}} = \frac{1}{C} + \frac{1}{K \cdot C} \]

### Step 4: Simplify the equation
To simplify the equation, find a common denominator:
\[ \frac{1}{C_{eff}} = \frac{K + 1}{K \cdot C} \]

### Step 5: Invert to find \( C_{eff} \)
Now, take the reciprocal to find \( C_{eff} \):
\[ C_{eff} = \frac{K \cdot C}{K + 1} \]

### Final Answer
Thus, the effective capacitance of the two capacitors connected in series, with one filled with a dielectric, is:
\[ C_{eff} = \frac{K \cdot C}{K + 1} \]