Question

When a 12V battery (i.e. a battery of EMF 12V) is connected across a lamp with a resistance of 6.8 Ω the PD across the lamp is 10.2V. Find (a) the current through the lamp, (b) the internal resistance of the battery. 

Answer 1

(a)

$$I=VR=10.2*6.8=69.4\:\rm A$$

(b)

$$r=\frac{{\cal E}-V}{I}=\frac{12-10.2}{69.3}=0.026\:\Omega$$