A battery whose e.m.f. is 6 V and whose internal resistance is 1.4Ω is applied to an external circuit consisting of two parallel resistances of 2 and 8Ω. Find the difference of potential at the terminals of the battery and the current in the resistances.
Answer: V = 3.2 V; I1 = 1.6 amperes; I2 = 0.4 ampere.
Explanation:
The resistance of the external circuit is
where r is the internal resistance of the battery. The voltage drop across the external circuit is
V = IR = 3.2V.
