Question

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 :

• A. 2.0 A
• B. 1.0 A
• C. 0.2 A
• D. 0.1 A

Answer 1

Answer: (C)

0.2 A

Answer 2

Total resistance of Wheatstone bridge $= \frac{(40)(120)}{40 + 120} = 30 \Omega$ Current through cell = $\frac{7 V}{( 5 +30) \Omega } = \frac{1}{5} A = 0.2 \, A$