Power dissipation in resistors in series versus in parallel

A student builds a circuit with a battery and two light bulbs in series. Then the student builds a second circuit with two light bulbs in parallel. Which battery runs out of power first?

Assume all bulbs have equal resistance. Assume both batteries have equal electrical potential. All right, so let's just think about this before we even look at the choices.

So the first scenario, you have your battery, positive terminal, negative terminal, and you have two light bulbs in series. So let's just draw it this way. Traditional light bulbs really are just resistors that when a current goes through them, they get hot enough, and that's because of the energy dissipated that they emit visible light. So that's the first resistor or the first light bulb, second one, and it goes back.

So that's the first scenario. The second scenario looks like this: plus minus where they are in parallel. They are in parallel just like that. And they say assume all bulbs have equal resistance. So we could just call their resistances R, just like that. We assume that both of these batteries are, for the sake of argument, identical.

So one way to think about it is what is going to be the equivalent resistance for these circuits? This one, when you have two resistors in series, you could simplify or you could view this circuit as you have your voltage source, which is our battery, and you could model it as a circuit with one resistor where its resistance is 2R.

If you wanted to think of an equivalent for this parallel circuit, you have your same voltage source, and you could model it as having a resistor there. What would this resistance be? What is equivalent to these two resistances in parallel?

So if we call this resistance sub 2, we know that 1 over resistance sub 2 is equal to 1 over R plus 1 over R. This is the formula for equivalent resistance when you have resistors in parallel. So this is going to be equal to 2 over R.

If you just take the reciprocal of this, you get R2 is equal to R over 2. So this is equal to R over 2. Now, what is the power dissipated across a resistor? We know, let me just do this in a different color. Power dissipated across a resistor is going to be the voltage across the resistor times the current going through that resistor.

Well, we know that the voltage across both of these resistances is going to be the same. So V is going to be the same in either situation. But what about the current? Well, we know that V is equal to IR. Another way to think about it is current is equal to voltage over resistance.

So our current in this first scenario is going to be V over 2R, while our current in this second scenario is going to be V over R over 2, which is the same thing as 2V over R. So our current in this second scenario is going to be 4 times as large.

So if you quadruple the current and you have the same voltage, well, you're going to quadruple the power. And so if you're quadrupling the power, you're quadrupling the amount of energy that's dissipated per unit time.

So in this second scenario, where we've quadrupled the power, we're quadrupling the energy dissipated over time. That battery is going to run out a lot faster.

So they ask which battery runs out of power first? Well, that's going to be the battery in the circuit with two light bulbs in parallel. And we're done.