To determine the frequency at which a capacitor of 2.5μF has a resistance of 250Ω in a radio circuit, we need to understand the relationship between capacitive reactance, resistance, and frequency. ### Step-by-Step Explanation 1. **Understanding Capacitive Reactance**: - The capacitive reactance (\(X_C\)) of a capacitor is given by the formula: \[ X_C = \frac{1}{2\pi f C} \] where: - \(X_C\) is the capacitive reactance in ohms (Ω), - \(f\) is the frequency in hertz (Hz), - \(C\) is the capacitance in farads (F). 2. **Given Values**: - Capacitance (\(C\)) = 2.5μF = \(2.5 \times 10^{-6}\) F - Resistance (\(R\)) = 250Ω 3. **Setting Up the Equation**: - In a circuit where a capacitor is used, the total impedance (\(Z\)) can be calculated using the resistance and the capacitive reactance. For a purely resistive-capacitive (RC) circuit, the impedance is given by: \[ Z = \sqrt{R^2 + X_C^2} \] - However, since we are looking for the frequency at which the capacitive reactance equals the resistance, we can set \(X_C = R\): \[ \frac{1}{2\pi f C} = R \] 4. **Rearranging the Formula**: - Rearranging the equation to solve for frequency (\(f\)): \[ 2\pi f C = \frac{1}{R} \] \[ f = \frac{1}{2\pi R C} \] 5. **Substituting the Values**: - Now, substituting the known values into the equation: \[ f = \frac{1}{2\pi \times 250 \times 2.5 \times 10^{-6}} \] - Calculating the denominator: \[ 2\pi \times 250 \times 2.5 \times 10^{-6} = 2\pi \times 0.000625 = 0.00392699 \text{ (approximately)} \] - Now, calculating the frequency: \[ f = \frac{1}{0.00392699} \approx 254.65 \text{ Hz} \] 6. **Finding the Correct Option**: - To express this frequency in terms of the options given, we can convert it to a form that matches the options: - We know that \(f = \frac{1}{2\pi \times 250 \times 2.5 \times 10^{-6}}\) can be simplified to: \[ f = \frac{1}{2\pi \times 250 \times 2.5 \times 10^{-6}} = \frac{1}{(2 \times 250 \times 2.5) \times \pi \times 10^{-6}} = \frac{1}{1250\pi \times 10^{-6}} = \frac{10^6}{1250\pi} = \frac{800}{\pi} \text{ Hz} \] - Therefore, the correct answer is **B. \( \frac{800}{\pi} \text{ Hz} \)**. ### Why Other Options Are Incorrect - **Option A: \( \frac{\pi}{800} \text{ Hz} \)**: - This option suggests a very low frequency, which does not match our calculations. The relationship between frequency and reactance indicates that higher capacitance or lower resistance would yield lower frequencies, but not in this case. - **Option C: \( 200\pi \text{ Hz} \)**: - This option suggests a much higher frequency than calculated. The capacitive reactance decreases with increasing frequency, and thus this option does not satisfy the condition of \(X_C = R\). - **Option D: \( 2000\pi \text{ Hz} \)**: - Similar to option C, this frequency is excessively high and does not align with the calculated frequency. ### Revision Summary - The frequency at which a capacitor has a specific resistance can be calculated using the formula \(f = \frac{1}{2\pi RC}\). - For a 2.5μF capacitor and 250Ω resistance, the frequency is \( \frac{800}{\pi} \text{ Hz} \). - Capacitive reactance decreases with increasing frequency, which is crucial for understanding RC circuits. - Always check the units and ensure they are consistent when performing calculations.