Answer to Question #263208 in Electrical Engineering for Neilmar

Question

A battery of 10V emf and internal resistance of 0.50Ω is connected in parallel with another battery of 12V and internal resistance of 20Ω. What is the load current?

Answer 1

The load current is

$$I=\frac{V_1-V_2}{r_1+r_2}=\frac{12-10}{20+0.5}=0.098\text{ A}.$$