A real battery is not just an emf we can model a real 15 vv...

A real battery is not just an emf. We can model a real 1.5 VV battery as a 1.5 VV emf in series with a resistor known as the "internal resistance", as shown in the figure(Figure 1). A typical battery has 1.0 ΩΩ internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 VV, the value of the emf. Suppose the terminals of this battery are connected to a 2.7 ΩΩ resistor. 1. What is the potential difference between the terminals of the battery? 2. What fraction of the battery's power is dissipated by the internal resistance?

Transcript

00:02 So here in this problem, we've got a battery 1 .5 volts with an internal resistance of 1 oom.

00:13 It's kind of inside the battery.

00:15 So we just put them in series like this because it's going through both of them and the terminals of the battery.

00:20 We'd say are on either side of that.

00:23 And then we've connected that to a 2 .7 oom resistor.

00:29 And the first question, ask what's the potential difference between the terminals of the battery? well, the terminal is the battery, you can think of as here and here, are also the voltage across the 2 .7 oom resistor across here.

00:45 So let's go ahead and first calculate the current for the whole circuit.

00:49 Okay.

00:50 I for the whole circuit is going to be the voltage divided by the resistance, which is 1 .5 volts divided by 3 .7 oms.

01:01 Because remember, to get the resistance of two resistors in series, which these are, it goes through one and then the other, you add them up.

01:09 Okay, and that gives us a current of 0 .405 amps.

01:21 Now, the next thing we need to do to get the voltage here is we need to focus in.

01:26 We'll call this resistor a just for a good measure.

01:29 Have a name for it.

01:30 The voltage across resistor a, which is also the voltage across the battery, we can take that current, the current of, sorry, to get the voltage, we get to want to do current, divided by the resistance.

01:51 And we want to do all this for light bulb or resistor a.

01:56 Okay.

01:57 So we plug in the current in a series circuit.

01:59 The current's the same everywhere.

02:01 So our current is 0 .1...

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