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An automobile starter motor is connected to a 12.0 $\mathrm{V}$ battery. When the starter is activated it draws 150 $\mathrm{A}$ of current, and the battery voltage drops to 7.0 $\mathrm{V} .$ What is the battery's internal resistance?
$$\begin{aligned} r &=\frac{12.0 \mathrm{V}-7.0 \mathrm{V}}{150 \mathrm{A}} \\ &=0.0333 \Omega \\ &=33.33 \mathrm{m} \Omega \end{aligned}$$
Physics 102 Electricity and Magnetism
Chapter 19
Current, Resistance, and Direct-Current Circuit
Electric Charge and Electric Field
Gauss's Law
Electric Potential
Capacitance and Dielectrics
Current, Resistance, and Electromotive Force
Direct-Current Circuits
Electromagnetic Induction
Cornell University
University of Michigan - Ann Arbor
University of Washington
Simon Fraser University
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So initially we have the publishers with some internal assistance. But the switch is open. There is no current flowing. So now when you consider the world is difference which is cm of minus I are There is no current current zero, which means this is simply the IMF on this we know is 12 old. Hence the IMF off This battery is full of old. Now that we have a lord connected Sistol world, the current the voltage across the battery has dropped to seven world instead. Now we Abie is excellent minus r and we know that the current given is 150 amperes and speak it. I'm sorry. This is excellent. We get from this equation The internal assistances Epsilon minus V be divided by which is 12 minus seven because the baby is sewn in this case seven words You have a current which is 1 50 happier. It starts out to be 0.33 This is the internal resistance off the battery
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