Question

In an LCR series circuit an alternating voltage source of frequency \(F\) is connected. The current leads the voltage by \(45^{\circ}\). The value of \(L\) is

• A. \(\frac{1+2πFCR}{2πFC}\)
• B. \(\frac{1+2πFCR}{4π^2F^2C}\)
• C. \(\frac{1-2πFCR}{4π^2F^2C}\)
• D. \(\frac{1+2πFR}{4π^2F^2C}\)

Hint:

Current leads the voltage when the circuit is capacitive; tan of the phase angle is (X_C - X_L) over R.

Answer 1

The Correct Option is C

Step 1: Approach
Write the impedance triangle.

Step 2: Triangle
The current leads by $45^\circ$, so the capacitive excess reactance equals the resistance: $X_C-X_L=R$.

Step 3: Rearrange
$X_L=X_C-R$, that is $2\pi FL=\dfrac{1}{2\pi FC}-R$. Multiply by $2\pi FC$: $4\pi^2F^2LC=1-2\pi FCR$.

Step 4: Answer
$L=\dfrac{1-2\pi FCR}{4\pi^2F^2C}$, option (C).

Final Answer:
Equal reactance difference and resistance give L = (1 - 2 pi F C R) over 4 pi squared F squared C, option (C). \[ \boxed{\frac{1-2\pi FCR}{4\pi^2F^2C}} \]