Question

A Wheatstone network ABCDA has the resistances 20 \(\Omega\), 10 \(\Omega\), and 12 \(\Omega\) in AB, BC, and DA arms respectively in which the galvanometer is connected across BD. For null deflection in the galvanometer, the resistance in CD arm should be:

• A. 5 \(\Omega\)
• B. 4 \(\Omega\)
• C. 10 \(\Omega\)
• D. 8 \(\Omega\)
• E. 6 \(\Omega\)

Hint:

In a Wheatstone bridge, for null deflection of the galvanometer, the ratio of the resistances in opposite arms must be equal.

Answer 1

Answer: (E)

In a Wheatstone bridge, for null deflection in the galvanometer, the following condition must be satisfied:
\[ \frac{R_{AB}}{R_{BC}} = \frac{R_{DA}}{R_{CD}} \] where: - \( R_{AB} = 20 \, \Omega \),
- \( R_{BC} = 10 \, \Omega \),
- \( R_{DA} = 12 \, \Omega \),
- \( R_{CD} \) is the unknown resistance we need to find.
Substitute the known values into the equation: \[ \frac{20}{10} = \frac{12}{R_{CD}} \] Simplifying the equation: \[ 2 = \frac{12}{R_{CD}} \] Solving for \( R_{CD} \): \[ R_{CD} = \frac{12}{2} = 6 \, \Omega \] Thus, the resistance in the CD arm should be \( 6 \, \Omega \). \[ \boxed{6 \, \Omega} \]