A battery of emf E produces currents I1 and I2 when connected to external resistance R1 and R2 respectively. The internal resistance of the battery is
plzzz explain.....
By Ohm's Law,
I1(R1+r)=E
and I2(R2+r)=E
Assuming r to be the internal resistance.
Equating the two equations, we get:
I1(R1+r)=I2(R2+r)
(I1R1-I2R2)/I2-I1)=r, internal resistance
I1(R1+r)=E
and I2(R2+r)=E
Assuming r to be the internal resistance.
Equating the two equations, we get:
I1(R1+r)=I2(R2+r)
(I1R1-I2R2)/I2-I1)=r, internal resistance