The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be
- 0.1 A
- 2.0 A
- 1.0 A
- 0.2 A
Answer
(4) 0.2 A
Solution
PQ = 1030 = 13 and RS = 3090 = 13
∵ PQ = RS, the bridge is balanced, so no current flows through the galvanometer and it can be removed.
Resistance of arm P + Q = 10 + 30 = 40 ; arm R + S = 30 + 90 = 120
These two are in parallel: Rp = 40 12040 + 120 = 30
Total resistance with the internal resistance = 30 + 5 = 35
∴ I = 735 = 0.2 A
Answer: (4)