Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 :

NEETNEET 2013Current Electricity

Solution:

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 $