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: Understanding the Concept
In a series LCR circuit the phase angle \(\phi\) satisfies \(\tan\phi=\dfrac{X_C-X_L}{R}\) when the current leads. The current leads by \(45^\circ\), so \(\tan45^\circ=1\).

Step 2: Equation
\[ X_C-X_L=R \]
\[ \frac{1}{2\pi FC}-2\pi FL=R \]

Step 3: Solve for L
\[ 2\pi FL=\frac1{2\pi FC}-R=\frac{1-2\pi FCR}{2\pi FC} \]
\[ L=\frac{1-2\pi FCR}{4\pi^2F^2C} \]

Step 4: Check the options
Options (B) and (D) have a plus sign, which would belong to a voltage-leading situation. Option (A) has the wrong dimension for an inductance. The answer is 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). \[ \boxed{\frac{1-2\pi FCR}{4\pi^2F^2C}} \]