Question

The resistances of the four arms P ,Q,R & 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. 0 . 2 A
• B. 0. 1 A
• C. 2 . 0 A
• D. 1 . 0 A

Answer 1

The correct option is A 0 . 2 A

$R_{net}=\frac{(30+10)\times (90+30)}{40+120}+5$

$=35\Omega$

$I=\frac{V}{R_{net}}$

$=\frac{7}{35}$

$=0.2A$

Option 'A'