Question #263208

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?

Expert's answer

The load current is


I=V1−V2r1+r2=12−1020+0.5=0.098 A.I=\frac{V_1-V_2}{r_1+r_2}=\frac{12-10}{20+0.5}=0.098\text{ A}.


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