Circuit analysis - two sources of emf

AI Thread Summary
The discussion revolves around a circuit with a 12 V battery and a 15 V charging unit, where confusion arises regarding the potential difference (p.d.) across the charging unit. The battery maintains a steady 12 V due to its negligible internal resistance, which means the charging unit's terminal p.d. must also be 12 V to avoid draining the battery. The difference between the charging unit's emf of 15 V and the terminal p.d. indicates that 3 V is lost across its internal resistance, resulting in a current of 6 A. The battery effectively powers the charging unit, ensuring the p.d. remains equal to its own. Understanding this relationship is crucial for analyzing circuits with multiple emf sources.
Miss_e101
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Hello all!

I'm confused about an explanation that my textbook gives, and I'm wondering if someone can help me understand it. The passage is as follows:

A circuit consists of a 12 V battery with negligible internal resistance, in parallel with a charging unit with an emf of 15 V and internal resistance of 0.5 ohms. These two components provide power for all the car's electrical equipment.

The car's battery maintains an emf of 12 V across it, because of its negligible internal resistance. Hence, the terminal p.d. of the charging unit must also be 12 V. This implies that 3 V must be "lost volts" across the internal resistance of the charging unit. Since its internal resistance is 0.5 V, there must be a current of 6 A supplied by the charging unit.


I'm confused specifically with the underlined statement - why does the p.d. of the charging unit have to be 12 V as well? Is it because the battery is actually powering the charging unit? Why then does it have an emf of 15 V?

Any insight would be appreciated.

Miss E.
 
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Welcome to PF!

Please take some time to read the site guidelines. We ask that all homework questions be formatted using the homework template so we can gauge your understanding.

With respect to your question, the charging unit is trying to charge the battery. A car battery is used primarily to start the car and after that the engine powers a generator to keep the battery charged and to supply the necessary amperage to power the cars electrical system. There should be no drain on the battery.
 
Last edited:

Homework Statement


Hello all!

I'm confused about an explanation that my textbook gives, and I'm wondering if someone can help me understand it. The passage is as follows:

A circuit consists of a 12 V battery with negligible internal resistance, in parallel with a charging unit with an emf of 15 V and internal resistance of 0.5 ohms. These two components provide power for all the car's electrical equipment.

The car's battery maintains an emf of 12 V across it, because of its negligible internal resistance. Hence, the terminal p.d. of the charging unit must also be 12 V. This implies that 3 V must be "lost volts" across the internal resistance of the charging unit. Since its internal resistance is 0.5 V, there must be a current of 6 A supplied by the charging unit.


I'm confused specifically with the underlined statement - why does the p.d. of the charging unit have to be 12 V as well? Is it because the battery is actually powering the charging unit? Why then does it have an emf of 15 V?

Any insight would be appreciated.

Miss E.

Homework Equations


Kirchoff's laws

The Attempt at a Solution


n/a
 
Thank you, I've moved the question to the proper forum.
 
Miss_e101 said:
why does the p.d. of the charging unit have to be 12 V as well?
Let the terminals of the charging unit be A and B, those of the battery be C and D, with A connected to C, B connected to D.
You know the potential difference between C and D, independently of current. You can take the resistances of AC and BD to be very low. If the p.d. between A and B exceeds that between C and D, what does that tell you about the currents in AC, BD?
 
The keyword here is "terminal". You have an ideal voltage source in the form of a battery, and whatever points it is connected across are forced to a P.D. of 12V. Being almost ideal, the battery can sink (and source) whatever current you ask of it, all the while maintaining that steady 12V across its terminals.

In practice you need to ensure the circuit it is connected to includes some resistance, otherwise there will be nothing to limit the battery's current to a safe, finite value.
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