In a potentiometer, a wire of length 10m having resistance `50 Omega` is used. A battery of 5V and a resistor of `450Omega` are connected in series to the wire. If an unknown battery of emf E balances the potentiometer at 450cm, then the value of E is

Options

A. 0.225V
B. 1.25V
C. 2.25 V
D. 0.0225V

Answer

A

Explanation

Given length of wire, l = 10m, resistance of wire, `R = 50 Omega`, emf of battery, `E_(1) = 5V`, balancing length, x = 450cm = 4.5m
series resistor, `R_(1) = 450 Omega`
Current, `i=(E_(1))/(R + R_(1)) = (5)/(50 + 450) = (5)/(500) = 0.01A`
So, `V=iR = 01 xx 50 = 0.5V`
emf of primary cell,
`E = (V_(x))/(l) = (0.5 xx 4.50)/(10) = 0.225V`