A capacitor is connected to an AC supply at 30 Hz. The reactance of the capacitor is 80 `Omega` . At what frequency its reactance will be 120 `Omega` ?

Answer

To solve the problem, we need to use the formula for capacitive reactance (Xc) which is given by:

\[ X_c = \frac{1}{2 \pi f C} \]

Where:
- \( X_c \) is the capacitive reactance in ohms (Ω),
- \( f \) is the frequency in hertz (Hz),
- \( C \) is the capacitance in farads (F).

### Step 1: Understand the relationship between frequency and reactance
The capacitive reactance \( X_c \) is inversely proportional to the frequency \( f \). This means that if the frequency increases, the reactance decreases, and vice versa.

### Step 2: Set up the known values
From the problem, we know:
- At frequency \( f = 30 \) Hz, the reactance \( X_c = 80 \) Ω.
- We want to find the new frequency \( f' \) at which the reactance \( X_c' = 120 \) Ω.

### Step 3: Use the relationship of reactance and frequency
Since \( X_c \) is inversely proportional to \( f \), we can write:

\[ \frac{X_c}{X_c'} = \frac{f'}{f} \]

Substituting the known values:

\[ \frac{80}{120} = \frac{f'}{30} \]

### Step 4: Solve for the new frequency \( f' \)
Cross-multiplying gives us:

\[ 80 \cdot 30 = 120 \cdot f' \]

Calculating the left side:

\[ 2400 = 120 \cdot f' \]

Now, divide both sides by 120 to find \( f' \):

\[ f' = \frac{2400}{120} = 20 \text{ Hz} \]

### Conclusion
Thus, the frequency at which the reactance will be 120 Ω is **20 Hz**.

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