Question

The resistances in the two arms of the meter bridge are 5Ω and RΩ, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 l₁. The resistance 'R', is:

• A. 15Ω
• B. 20Ω
• C. 25Ω
• D. 10Ω

Answer 1

The Correct Option is A

The problem involves a meter bridge circuit where you need to find the resistance R using the concept of shunt. Initially, we have two arms with resistances 5Ω and R \Omega. Upon shunting R with an equal resistance, the new balance point is at 1.6 l_1.

1. Initially, let the balance point be at a length l_1.
2. According to the principle of a meter bridge (Wheatstone bridge), we have: \frac{5}{R} = \frac{l_1}{100 - l_1}.
3. When the resistance R is shunted with an equal resistance, the effective resistance becomes: \frac{R \times R}{R + R} = \frac{R}{2}.
4. Now, the balance point shifts to 1.6 l_1. Use the new balance condition: \frac{5}{R/2} = \frac{1.6l_1}{100 - 1.6l_1}.
5. Simplifying the equation: \frac{10}{R} = \frac{1.6l_1}{100 - 1.6l_1}.
6. We know from the initial condition: \frac{5}{R} = \frac{l_1}{100 - l_1}. Substitute the second condition: \frac{10}{R} = \frac{1.6l_1}{100 - 1.6l_1} \Rightarrow 10 \times (100 - 1.6l_1) = R \times 1.6l_1.
7. Using the two equations: \frac{1}{2} = \frac{1.6l_1 \times (100 - l_1)}{l_1 \times (100 - 1.6l_1)}.
8. By solving the above relation, the value of R can be determined: R = 15 \Omega.

Thus, the correct answer is 15Ω.