In the circuit given below a current of 0.8 A flows through the external resistance when R= 10`Omega` and 0.4A when R = 25.`Omega`. Calculate the internal resistance and e.m.f. of the battery.

Explanation

Current `i=(R)/(R+r) `
EMF of the battery E=i(R+r)
Hence 0.8=`(E)/(10+r)"------"(1)`
`0.4=(E)/(25+r)"-----"(2)`
eq. (1) eq .`(2)rArr(0.8)/(0.4)=(25+r)/(10+r)`
Solving we get `r=5Omega " " therefore`E=0.8(10+5)=12V
Thus the battery has an e.m.f of 12 V and an internal resistance of 5`Omega`