Internal resistance of a battery and its affect on power drain TBR

[javksmith93]

Hey all TBR question electric circuits question 10 page 203

How does the internal resistance of a battery, Rbattery, affect the power drain by R1 and R2?
A. Both drain less power.
B. Both drain more power.
C. The power drain does not change in either resistor.
D. It depends upon how big Rbattery is compared to Rl and R2.

We are talking about a series circuit that now includes an extra resistor, Rbattery- You can imagine that the more resistors you have in series, the harder it is for the battery to push a current. Thus, this's extra resistance results in a slightly smaller current. The power drained by the resistors is:
P=I2R
Thus, if the current, I, decreases a bit, then so will the powered drained. Note that R in the equation does not change ,because it is a property of the resistor itself, and not dependent on the circuit in which it is placed. The best answer is choice A.

I don't understand the explanation the circuits are as the explanation says in series so shouldn't the current I be the same over all of them?

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[sazerac]

The internal resistance of the battery is just another resistor in series. So if this new resistor gets larger, the resistance of the whole circuit gets larger, and the current in the whole circuit goes down.

 

[StarFall]

The internal resistance of the battery is just another resistor in series. So if this new resistor gets larger, the resistance of the whole circuit gets larger, and the current in the whole circuit goes down.

How does this relate with power drain? I'm trying to understand what 'power drain' means from a conceptual standpoint.
I understand that adding a battery with some resistance (thus it's just another resistor) to the circuit increases the total resistance in the circuit. Increase resistance leads to decrease current.
Since P = IV, voltage is always held constant (?) and since the current in the whole circuit decreased, the power or power drain will decrease?

Or is it like when you add a resistor/increase the resistance of the circuit, the resistors R1 and R2 will have a smaller current. Since they have a smaller current, the voltage drop across R1 and R2 are now decreased since Rbattery can now 'help out' in decreasing the voltage drop across the whole circuit. So R1 and R2 will drain less voltage/power with Rbattery than R1 and R2 alone. Am I on the right track?

 

[sazerac]

Yes you are on the right track. P=IV, and we already proved I will be smaller. As you surmised, the V across the old resistors will also be lower. So for sure the P=IV across each resistor will be lower.

Sometimes it helps to think in terms of extremes. Let's say the battery resistance was humongous. So big it basically cut the wire in the circuit. What would be the power consumption then?

 

[NextStepTutor_1]

Yes, go to large extremes with these questions. I completely agree w/ @sazerac (love the name).
Don't be afraid to push the scenario to "wild" extremes when the question introduces a change in the system.

Adding a resistor in series increases the total resistance of the circuit, and if the nature of the battery isn't changing outside of the it's internal resistance is increasing, then the voltage provided by the battery should not be changing (though the voltage across any one of the resistors will now decrease since there will be less current). Using, V=IR, we thus see that the current available to the circuit must decrease and using P=IV, as current goes down, so too does power.

 

[StarFall]

Yes you are on the right track. P=IV, and we already proved I will be smaller. As you surmised, the V across the old resistors will also be lower. So for sure the P=IV across each resistor will be lower.

Sometimes it helps to think in terms of extremes. Let's say the battery resistance was humongous. So big it basically cut the wire in the circuit. What would be the power consumption then?

With a huge resistance, the power consumption would be all at that battery with R1 and R2 with about 0 power consumption.
Rbattery would essentially stop the current, almost no e- can pass through the resistor, so the voltage drop would be about equal to total voltage we began with. The 'leakage' of current/e- flow would be split between R1 and R2, and they would essentially have no voltage drop.

@sazerac, thanks so much for the help! @NextStepTutor_1 thank you for your clarification. Using the simple equations of V=IR and P=IV really helped to nail this concept down for me.
I hope OP sees these posts and helps to hit the nail on the head for him/her.