Question

for a cell potential difference is 3.6v, when the circuit is open . if the potential difference reduces to 3v , when cell is connectedto a resistance to 5 ohm ,the internal resistance of cell is

Answer 1

Explanation:

E=i(R+r)

i=E/(R+r)

i=3.6/5+r

potential drop =i×R

3=[3.6/5+r]×5

5+r=6

r=1 ohm

Answer 2

Answer:

r = 0.6 ohms

Explanation:

P.D = 3.6 volts , this is the EMF of the cell.

Now when the circuit is closed and a current is flowing through the circuit then the P,D = 3 v

Thus the total resistance of the circuit is R +r= 5 + r ohms

 so , the current in the circuit is i = 3.6/5+r

hence the P>D across the internal resistance .

Given that,

r(3.6/5+r)=3.6-3 as the drop in the potential across the battery is due the P.D across the internal resistance

so, r(3.6/5+r) = 0.72

3r= 3.6

r=3.6-3

r=0.6 ohms