A series circuit consist of 6 ohms resistor,ammeter and battery. The current in the circuit is 0.5A but drops to 0.3A when another resistor of 8ohms is added in series. Find the battery e.m.f and its internal resistance?

Answer 1

E = 0.5*r + 0.5*6
Eq1: E = 0.5r + 3.

E = 0.3*r + 0.3*(6+8).
Eq2: E = 0.3r + 4.2.

E = 0.5r + 3 = 0.3r + 4.2.
0.5r +3 = 0.3r + 4.2.
0.2r = 1.2.
r = 6 Ohms = Internal resistance.

E=0.5r + 0.5*R = 0.5*6 + 0.5*6 = 6 Volts