Understanding Current in Series Circuits: Explaining 6A with 12V and 2 Ohms

In summary: In the figure, the voltage across the resistor X is changing as a result of the current flowing through it.
  • #1
Lim Y K
26
0
Take this question as an example
ImageUploadedByPhysics Forums1440066974.082733.jpg

They asked for the current flowing through the bulb and the answer is 6A
I know that it's 12V divided by 2 ohms(2 ohm is the resistance of resistor X) , but why? I thought the current is supposed to be 12V divided by the resistance of bulb and resistor X(total resistance of resistors) since its arranged in series?
 
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  • #2
As stated, the voltage across the resistor X is measured at 12 V, so the other parts of the circuit shouldn't matter.
 
  • #3
Yes, current would be 12/(rtotal), where rtotal is the sum of the series resistances.

I'm not able to agree or disagree with you on the correctness of the textbook's answer, because you haven't disclosed the exact wording of the question. The figure you selected to use as an example is not a particularly good choice---because it examplifies a fundamental wiring "mistake".
 
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Likes CWatters and davenn
  • #4
olivermsun said:
As stated, the voltage across the resistor X is measured at 12 V, so the other parts of the circuit shouldn't matter.

for a idea circuit, yes you could say that. But if it was a real world circuit, that is no longer true. The light globe will have significant resistance
which will change as it heats up

NascentOxygen said:
The figure you selected to use as an example is not a particularly good choice---because it examplifies a fundamental wiring "mistake".

indeed

@Lim Y K, can you recognise the problem ?Dave
 
  • #5
davenn said:
for a idea circuit, yes you could say that. But if it was a real world circuit, that is no longer true. The light globe will have significant resistance
which will change as it heats up
I realize that is the case for a real incandescent lightbulb, but since the heated filament resistance, wasn't given in the problem, I'm assuming that the student isn't expected to incorporate that information in the solution.
 
  • #6
maybe but they need to be informed that it isn't like that in the real world and you didn't give that clarification :smile:

the poor guy is already getting confused enough with incorrectly drawn circuits ...
 
  • #7
olivermsun said:
As stated, the voltage across the resistor X is measured at 12 V, so the other parts of the circuit shouldn't matter.
This is not the case in the figure I'm seeing.
 

1. What is current in series?

Current in series is the flow of electrical charge through a series circuit, where all components are connected end-to-end with no branching paths. It is the same at all points in the circuit and is determined by the voltage of the power source and the resistance of the components.

2. How is current in series calculated?

To calculate the current in a series circuit, you can use Ohm's law, which states that current (I) is equal to voltage (V) divided by resistance (R). This can also be represented by the equation I = V/R.

3. What happens to current in series when resistance is increased?

When resistance in a series circuit is increased, the current will decrease. This is because there is more resistance for the current to overcome, causing it to flow at a slower rate.

4. How does current in series differ from current in parallel?

In a series circuit, the current is the same at all points, while in a parallel circuit, the current is divided between multiple paths. Additionally, in a series circuit, the total resistance is equal to the sum of the individual resistances, whereas in a parallel circuit, the total resistance is less than any of the individual resistances.

5. Can current in series ever be greater than the current of the power source?

No, in a series circuit, the current is limited by the resistance of the components and cannot be greater than the current of the power source. In fact, the total resistance in a series circuit will always limit the amount of current that can flow through the circuit.

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