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A battery having an emf of 9.00 $V$ delivers 117 $\mathrm{mA}$ when connected to a $72.0-\Omega$ load. Determine the internal resistance of the battery.

$$

R_{\text {internal}}=4.92 \Omega

$$

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University of Michigan - Ann Arbor

University of Washington

Hope College

University of Sheffield

So you're given that the voltage is equal to nine and that the current is equal to 117 million amps, which if you convert that two amps, is going to be equal to 0.117 amps Then you're also told that our load or the lone resistance is equal to 72 homes. So here we can just use homes Law, which is going to be V equals I r Ah. In this case are means the total resistance, so we can just plug in the numbers that we know. So we know that voltage is equal to nine. We know that the current is equal to 0.11 amperes. So then we can just find our total. In this case, our total is going to be equal. Thio 76 0.9 to 3. So to calculate the internal resistance, we can think about what our total is. So our total is going to be the same as saying our load plus our internal. So if we just rearrange that, then our internal would be equal Thio our total minus are load And since we know both of those numbers, we can just plug it in so it would be equal to 76 0.9 to 3. Owns minus 72 homes. So doing that, you get that the internal resistance is equal to 4.9 to 3. And don't forget units, homes.

University of California - Los Angeles