Question

In Maxwell's capacitance bridge for calculating unknown inductance, the various values at balance are \( R_2 = 300 \, \Omega \), \( R_3 = 700 \, \Omega \), \( R_4 = 1500 \, \Omega \), \( C_4 = 0.8 \, \mu\text{F} \). Calculate \( R_1 \), \( L_1 \), and Q factor, if the frequency is 1100 Hz.

• A. \( 240 \, \Omega,\ 0.12 \, \text{H},\ 3.14 \)
• B. \( 140 \, \Omega,\ 0.168 \, \text{H},\ 8.29 \)
• C. \( 140 \, \Omega,\ 0.12 \, \text{H},\ 5.92 \)
• D. \( 240 \, \Omega,\ 0.36 \, \text{H},\ 8.29 \)

Hint:

In Maxwell's capacitance bridge, the unknown values can be directly calculated using: \[ R_1 = \frac{R_2 R_3}{R_4}, \quad L_1 = C_4 R_2 R_3 \] Always remember to convert microfarads to farads. To find the Q factor, use \( Q = \frac{\omega L_1}{R_1} \) with \( \omega = 2\pi f \).

Answer 1

The Correct Option is B

Maxwell's capacitance bridge is used to measure an unknown inductance \( L_1 \) and its associated resistance \( R_1 \) by comparing it with a standard capacitor and resistors. At balance, the condition is: \[ Z_1 Z_4 = Z_2 Z_3 \] Where:

• \( Z_1 = R_1 + j\omega L_1 \)
• \( Z_2 = R_2 \)
• \( Z_3 = R_3 \)
• \( Z_4 = \frac{R_4}{1 + j\omega C_4 R_4} \)

Substitute into the balance equation: \[ (R_1 + j\omega L_1) \cdot \frac{R_4}{1 + j\omega C_4 R_4} = R_2 R_3 \] Multiplying: \[ (R_1 + j\omega L_1) R_4 = R_2 R_3 (1 + j\omega C_4 R_4) \] Separate real and imaginary parts: Real part: \[ R_1 R_4 = R_2 R_3 \Rightarrow R_1 = \frac{R_2 R_3}{R_4} \] Imaginary part: \[ \omega L_1 R_4 = \omega C_4 R_4 R_2 R_3 \Rightarrow L_1 = C_4 R_2 R_3 \] Given: \begin{align*} R_2 &= 300 \, \Omega \\ R_3 &= 700 \, \Omega \\ R_4 &= 1500 \, \Omega \\ C_4 &= 0.8 \, \mu\text{F} = 0.8 \times 10^{-6} \, \text{F} \\ f &= 1100 \, \text{Hz} \end{align*} 1. Calculate \( R_1 \): \[ R_1 = \frac{300 \times 700}{1500} = \frac{210000}{1500} = 140 \, \Omega \] 2. Calculate \( L_1 \): \[ L_1 = 0.8 \times 10^{-6} \times 300 \times 700 = 0.168 \, \text{H} \] 3. Calculate Q factor: \[ Q = \frac{\omega L_1}{R_1}, \quad \omega = 2\pi f = 2\pi \times 1100 \approx 6911.5 \, \text{rad/s} \] \[ Q = \frac{6911.5 \times 0.168}{140} \approx \frac{1161.132}{140} \approx 8.29 \] Final Results: \[ \boxed{R_1 = 140 \, \Omega,\quad L_1 = 0.168 \, \text{H},\quad Q = 8.29} \]