Circuit analysis - two sources of emf

In summary: The resistance of the circuit will then limit the maximum current that can flow from the battery to the charging unit.
  • #1
Miss_e101
4
0
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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  • #2
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:
  • #3

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
 
  • #4
Thank you, I've moved the question to the proper forum.
 
  • #5
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?
 
  • #6
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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Related to Circuit analysis - two sources of emf

1. What is circuit analysis and why is it important?

Circuit analysis is the process of studying and understanding the behavior and characteristics of electrical circuits. It is important because it allows us to design and troubleshoot circuits, which are the backbone of modern technology.

2. What are the two sources of emf in a circuit?

The two sources of emf (electromotive force) in a circuit are voltage sources and current sources. Voltage sources, such as batteries or generators, provide a constant potential difference between two points in a circuit. Current sources, such as transistors or diodes, provide a constant flow of electrons through a circuit.

3. How do you analyze a circuit with two sources of emf?

To analyze a circuit with two sources of emf, you can use Kirchhoff's laws. Kirchhoff's voltage law states that the sum of all voltage drops in a closed loop must equal the sum of all voltage rises. Kirchhoff's current law states that the sum of all currents entering a node must equal the sum of all currents leaving the node. By applying these laws and using other circuit analysis techniques, you can determine the behavior and characteristics of the circuit.

4. What is the difference between series and parallel circuits?

In a series circuit, all components are connected in a single loop, so the current is the same at every point in the circuit. In a parallel circuit, the components are connected in multiple branches, so the current is divided among the branches. Additionally, the total resistance in a series circuit is the sum of the individual resistances, while the total resistance in a parallel circuit is less than any individual resistance.

5. How can you determine the voltage and current in a circuit with two sources of emf?

To determine the voltage and current in a circuit with two sources of emf, you can use Ohm's law and the equations for series and parallel circuits. Ohm's law states that voltage is equal to current multiplied by resistance (V=IR). For series circuits, you can use the equation V=IR to determine the voltage drop across each component. For parallel circuits, you can use the equation I=V/R to determine the current through each branch.

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