00:01
From reading the problem, we know that the battery is 15 volts.
00:07
The resistance of the appliance is 75 oms, and there is an ideal voltmeter that's connected across the appliance, which reads 11 .9 volts.
00:19
And in part a, we're asked to find how much power is being dissipated by the appliance.
00:29
Actually, we can readily find this if we use the equation, power is equal to v squared over r.
00:44
And this v squared is what the voltmeter reads.
00:52
So that is 11 .9 volt squared over 75 oms, which gives us a dissipated power of 1 .89 watts.
01:15
And now in part b, we're asked, what is the internal resistance of the battery? so in order to find the internal resistance of the battery, we just take the sum of the potential difference in a closed loop, right? so if we start at some point and we end there, then we know that the total potential difference is equal to zero.
01:48
So we have a battery, and then we have a bolted drop associated with the internal resistance of the battery, and then another voltage drop that's associated with the appliance, and that's, these are all the, uh, the bolted drops there, are the potential differences there in the circuit...