Batteries, Resistance, Power, Potential, and Current

AI Thread Summary
A 9V battery with an internal resistance of 1.5 ohms and a 3-ohm load connected will result in a current of 2 amps flowing through the circuit. The potential difference read by an ideal voltmeter across the battery's terminals with the load connected will be 6V. The confusion arises from understanding why the current through the load is the same as that through the battery, which is due to the series connection of components. Drawing a circuit diagram can clarify the relationship between the battery and the load, highlighting that the current remains constant in a series circuit. Understanding these principles is crucial for solving similar electrical problems effectively.
NkaujHelp
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Homework Statement


You have a 9V battery with an internal resistance of 1.5 ohms. If a 3 ohms load is connected to the battery, how much current will flow through it? When the 3 ohms load is connected, what potential difference will an ideal voltmeter read across the battery's terminals?

Homework Equations


V=IR

The Attempt at a Solution


Okay, for the first problem, I used the above equation for the battery and for the resistance load and set them equal to each other. So the equation for the battery from above is I=V/R, where V= 9V and R=1.5 ohms. The equation for the resistance load is V=IR, where R=3 ohms. Taking both equations, I got V/R=IR and solved for I, the current. The current came out to be 2 amps. What I don't understand from this problem is why I had to set those equations equal to each other to get the current for the resistance load. Why can't I do I=V/R, where V = 9V and R = 1.5 ohms. If I did that, is that the current through the battery itself? I don't get why the current for the resistance load is the same as for the battery.

For the second problem, I used V=IR again where I took the current from above, 2 A, and plugged it into that equation. V=(2A)(3ohms)=6 V. That is the potential difference an ideal voltmeter will read across the battery's terminals when the 3 ohms load is connected. I don't understand why I am using the 3 ohms load and its current to find the potential difference across a battery's terminals? I don't understand why the battery's own current and its own resistance, 1.5 ohms, was not used to solve this problem. Wouldn't the 6V be the potential difference across the resistance load itself?

I'm just having a hard time understanding the story behind the equations and what I did in these problems. Can someone please explain?
 
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NkaujHelp said:
Okay, for the first problem, I used the above equation for the battery and for the resistance load and set them equal to each other.
Can you be more specific about what you set equal to what and why? I suggest that you draw a circuit diagram first and study it before calculating anything.
 
Last edited:
NkaujHelp said:

Homework Statement


You have a 9V battery with an internal resistance of 1.5 ohms. If a 3 ohms load is connected to the battery, how much current will flow through it? When the 3 ohms load is connected, what potential difference will an ideal voltmeter read across the battery's terminals?

Homework Equations


V=IR

The Attempt at a Solution

What I don't understand from this problem is why I had to set those equations equal to each other to get the current for the resistance load. Why can't I do I=V/R, where V = 9V and R = 1.5 ohms. If I did that, is that the current through the battery itself? I don't get why the current for the resistance load is the same as for the battery.

It would help you if you drew the circuit. Ask yourself ... are the components in series or parallel with each other? and does that mean the current through each is the same or different?
 
NkaujHelp said:
Why can't I do I=V/R, where V = 9V and R = 1.5 ohms. If I did that, is that the current through the battery itself? I don't get why the current for the resistance load is the same as for the battery.
If you short the battery terminals together, what would be the current?
 
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