Question

when a resistance of 2 ohms is connected across the terminals of a cell the current is 0.5 amperes,when the resistance is increased to 5 ohms, the current is 0.25 amperes. the internal resistance of the cell is

Answer 1

When a resistance of 2Ω is connected across the terminals of a cell, the current is 0.5 A. When resistance is increased to 5Ω, the current is 0.25 A.

The emf of the cell is V= it, v= .5*(r+2)= (1/2)*(r+2), when resistance is increased to 5, then v= .25*(r+5)= (1/4)*(r+5), from above equations, (1/2)*(r+2)= (1/4)*(r+5), 2r+4= r+5, r= 1 ohm

Answer 2

emf=I(R+r) ---(a)

When I=0.5,R=2ohm

we know that,

emf=V

Therefore Eq(a)become V=0.5(2+r)=>1/2(2+r)

2V=2+r - - - (1)

When I=0. 25,R=5ohm

Therefore Eq(a)become

V=0.25(5+r)=>1/4(5+r)

4V=5+r - - - (2)

Subtract eq(1) from (2)

4V-2V=5+r-2-r

2V=3

V=3/2

V=1.5v (answer)