To find the inductance value that will give a resonance frequency of 200 kHz in a series LC circuit (which consists of a capacitor and an inductor), we can use the formula for the resonance frequency \( f_0 \): \[ f_0 = \frac{1}{2\pi\sqrt{LC}} \] Where: - \( f_0 \) is the resonance frequency in hertz (Hz), - \( L \) is the inductance in henries (H), - \( C \) is the capacitance in farads (F). ### Step 1: Rearranging the Formula We need to rearrange the formula to solve for \( L \): \[ L = \frac{1}{(2\pi f_0)^2 C} \] ### Step 2: Substituting the Values Given: - \( f_0 = 200 \, \text{kHz} = 200 \times 10^3 \, \text{Hz} \) - \( C = 20 \times 10^{-12} \, \text{F} \) Now, substituting these values into the formula: \[ L = \frac{1}{(2\pi (200 \times 10^3))^2 (20 \times 10^{-12})} \] ### Step 3: Calculating the Denominator First, calculate \( 2\pi (200 \times 10^3) \): \[ 2\pi (200 \times 10^3) \approx 1256.64 \, \text{Hz} \] Now, square this value: \[ (1256.64)^2 \approx 158,0000 \, \text{Hz}^2 \] Now, multiply this by the capacitance \( C \): \[ 1580000 \times (20 \times 10^{-12}) = 1580000 \times 20 \times 10^{-12} = 31.6 \times 10^{-6} = 3.16 \times 10^{-5} \] ### Step 4: Final Calculation for Inductance Now, take the reciprocal to find \( L \): \[ L = \frac{1}{3.16 \times 10^{-5}} \approx 31.64 \times 10^{4} \, \text{H} = 0.3164 \, \text{H} \] ### Step 5: Converting to the Given Options The options provided are in the form of \( I/n \) H. To express \( 0.3164 \, \text{H} \) in this form, we can see that: - \( 0.3164 \, \text{H} \) can be approximated as \( \frac{1}{32} \, \text{H} \) (since \( 1/32 \approx 0.03125 \, \text{H} \)). Thus, the correct answer is: **D. \( I/32 \, \text{H} \)** ### Explanation of Other Options - **A. \( I/16 \, \text{H} \)**: This would correspond to a much higher inductance value than what we calculated, leading to a lower resonance frequency. - **B. \( I/8 \, \text{H} \)**: Similar to option A, this would also yield a resonance frequency that is too low. - **C. \( I/64 \, \text{H} \)**: This would correspond to a very low inductance, resulting in a much higher resonance frequency than 200 kHz. ### Common Pitfalls - **Units**: Always ensure that the units are consistent (e.g., converting kHz to Hz). - **Calculating \( 2\pi f \)**: Make sure to calculate this correctly as it is crucial for finding the correct inductance. - **Rounding**: Be careful with rounding during calculations; it can lead to significant errors in the final answer. ### Revision Summary - The resonance frequency formula for an LC circuit is \( f_0 = \frac{1}{2\pi\sqrt{LC}} \). - Rearranging gives \( L = \frac{1}{(2\pi f_0)^2 C} \). - Substitute the values for \( f_0 \) and \( C \) to find \( L \). - The correct inductance for a resonance frequency of 200 kHz with a 20 pF capacitor is \( I/32 \, \text{H} \).