A 12V battery has an internal resistance of 0.05Ω. Two loads are connected in parallel across its terminal, the first drawing a current of 12A and the other taking a power of 200 watts. Determine the load terminals voltage?
"I=E\/(R+r)\\to IR+0.05I=12\\to"
"U+0.05I=12"
"I=I_1+I_2=12+200\/U" .So, we have
"U+(12+200\/U)\\cdot0.05=12"
"U^2-11.4U+10=0\\to U=10.4\\ (V)" . Answer
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