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 the above

Explanation

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

### Step 1: Identify the capacitance of the capacitors
Let the capacitance of the first capacitor (without dielectric) be \( C_1 = C \) and the capacitance of the second capacitor (with dielectric) be \( C_2 \). The capacitance of a capacitor filled with a dielectric is given by:

\[ C_2 = K \cdot C \]

### Step 2: Write the formula for effective capacitance in series
For two capacitors in series, the formula for the effective capacitance \( C_{eff} \) is given by:

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

### Step 3: Substitute the values of \( C_1 \) and \( C_2 \)
Substituting \( C_1 \) and \( C_2 \) into the formula:

\[ \frac{1}{C_{eff}} = \frac{1}{C} + \frac{1}{K \cdot C} \]

### Step 4: Simplify the equation
To simplify the right-hand side, we can find a common denominator:

\[ \frac{1}{C_{eff}} = \frac{K + 1}{K \cdot C} \]

### Step 5: Invert to find \( C_{eff} \)
Now, we can invert both sides to solve for \( C_{eff} \):

\[ C_{eff} = \frac{K \cdot C}{K + 1} \]

### Final Answer
Thus, the effective capacitance of the two capacitors in series, with one filled with a dielectric substance, is:

\[ C_{eff} = \frac{K \cdot C}{K + 1} \]

---