Question

An electronic device operates at $2\text{ MHz}$. The oscillating circuit has an inductance $20\times10^{-5}\text{ H}$. What is the capacitive reactance of the resonant circuit?

• A. $2512\text{ }\Omega$
• B. $1256\text{ }\Omega$
• C. $5024\text{ }\Omega$
• D. $251.2\text{ }\Omega$

Hint:

The keyword "resonant circuit" tells you everything you need to know. Do not waste time using the complex $X_C = \frac{1}{2\pi f C}$ formula to calculate capacitance first. Simply find $X_L$ since $X_C$ must equal $X_L$ at resonance!

Answer 1

The Correct Option is A

Concept: For an electronic tuned circuit to settle into electrical resonance at an operating frequency $f$, the inductive reactance ($X_L$) of the circuit loop must be exactly equal to its capacitive reactance ($X_C$): \[ X_C = X_L \] The inductive reactance of an inductor is calculated using the angular frequency relationship: \[ X_L = 2\pi f L \]

Step 1: Identify parameters and calculate the matching inductive reactance.
We are given:
• Operating resonance frequency, $f = 2\text{ MHz} = 2 \times 10^{6}\text{ Hz}$
• Circuit inductance, $L = 20\times10^{-5}\text{ H}$ Let's compute the value of $X_L$: \[ X_L = 2 \times \pi \times (2 \times 10^{6}) \times (20\times10^{-5}) \] \[ X_L = 2 \times \pi \times 2 \times 20 \times 10^{6 - 5} = 80\pi \times 10^1 = 800\pi\text{ }\Omega \]

Step 2: Apply the resonance condition to find the capacitive reactance.
Taking $\pi \approx 3.1416$: \[ X_L = 800 \times 3.14159 \approx 2513.27\text{ }\Omega \] Since the circuit is operating at resonance, $X_C = X_L$. Looking at the provided options, this rounds perfectly to: \[ X_C = 2512\text{ }\Omega \]