If the potential difference between B and D is zero, the value of x is 1/n Ω. The value of n is __________.

Solution

Physics sir · 5 steps

Collapse each arm to one resistor first, then the bridge is just a balanced Wheatstone ratio.

1. Upper arms. On the left, 6 Ω and 3 Ω are in parallel: (6 × 3)/(6+3) = 2 Ω. On the right, x and 1 Ω are in parallel: x/(1+x).
2. Lower arms. On the left, 1 Ω and 2 Ω are in series, giving 3 Ω. On the right there is a single x.
3. Balance condition. Zero potential difference between B and D means the bridge is balanced. The ratio of the two left arms equals the ratio of the two right arms.

   2/3 = (x/(1+x))/x = 1/(1+x)

4. Solve. 2(1+x) = 3, so x = 1/2 Ω. Check: the upper right arm is then 1/3 Ω, and 2/(1/3) = 3/(1/2) = 6 on both sides.
5. So. x = 1/2 Ω = 1/n Ω, so n = 2.

The idea to keep. When an arm of a bridge holds several resistors, reduce each arm to one equivalent resistance first. Then apply P/Q = R/S, keeping arms on the same side of the galvanometer together.

Answer from the NTA final answer key. Solution by PrepWiser.