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

Understanding the Problem:

• Maxwell's capacitance bridge is used to measure unknown inductance (\(L_1\)) and resistance (\(R_1\)).
• At balance, the impedance ratios of the bridge arms are equal.
• The Q factor (Quality Factor) of the coil is calculated using the formula \(Q = \frac{\omega L_1}{R_1}\).

Formulae:

• Balance condition for Maxwell's bridge: \(\frac{R_1}{R_2} = \frac{R_3}{R_4} + j\omega C_4 R_3\).
• Inductance calculation: \(L_1 = R_2 R_3 C_4\).
• Q factor: \(Q = \frac{\omega L_1}{R_1} = \omega C_4 R_4\).

Given Values:

• \(R_2 = 300 \, \Omega\)
• \(R_3 = 700 \, \Omega\)
• \(R_4 = 1500 \, \Omega\)
• \(C_4 = 0.8 \, \mu\text{F} = 0.8 \times 10^{-6} \, \text{F}\)
• Frequency (\(f\)) = 1100 Hz
• Angular frequency (\(\omega\)) = \(2\pi f = 2\pi \times 1100 \, \text{rad/s}\)

Calculating \(R_1\):

From the balance condition:

\(\frac{R_1}{R_2} = \frac{R_3}{R_4}\)

\(R_1 = R_2 \times \frac{R_3}{R_4}\)

\(R_1 = 300 \times \frac{700}{1500} = 140 \, \Omega\)

Calculating \(L_1\):

Using the inductance formula:

\(L_1 = R_2 R_3 C_4\)

\(L_1 = 300 \times 700 \times 0.8 \times 10^{-6}\)

\(L_1 = 168 \times 10^{-3} \, \text{H} = 0.168 \, \text{H}\)

Calculating Q Factor:

Using the Q factor formula:

\(Q = \omega C_4 R_4\)

\(Q = 2\pi \times 1100 \times 0.8 \times 10^{-6} \times 1500\)

\(Q = 2\pi \times 1100 \times 1.2 \times 10^{-3}\)

\(Q = 8.29\)

Final Answer:

\(R_1 = 140 \, \Omega\), \(L_1 = 0.168 \, \text{H}\), \(Q = 8.29\)