Question
A 25.0-$\Omega$ bulb is connected across the terminals of a 12.0-V battery having 3.50$\Omega$ of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?
Step 1
This is the sum of the resistance of the bulb and the internal resistance of the battery. So, $R_{total} = R_{bulb} + R_{internal} = 25.0\, \Omega + 3.50\, \Omega = 28.5\, \Omega$. Show more…
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