Parallel Circuit: Current with 4 vs 3 Resistors

AI Thread Summary
In a parallel circuit with four identical resistors, each carrying a current of 12 A, the total current is 48 A. When one resistor burns out, the remaining three resistors still carry 12 A each, but the total current drops to 36 A due to the increased equivalent resistance. The voltage across each resistor remains unchanged, which is why the current through each of the remaining resistors does not change. The confusion arises from the misunderstanding that the overall circuit still functions the same despite the loss of one resistor. Ultimately, the circuit's total current decreases when resistors are removed, even though individual currents remain constant.
Durfys
Messages
2
Reaction score
0
Hey I came across this question in my textbook and the answer provided just does't make sense.

A circuit consists of a battery and 4 identical resistors in parallel. The current in each resistor is 12 A. If one of the resistors burns out, what will be the current in each of the remaining 3 resistors?

Book's explanation:
The battery remains the same and will provide the same voltage both with the 4 resistors and with the 3 resistors. Since the resistors are in parallel, the voltage drop on each must be the same as the voltage provided by the battery, and will be unchanged when one resistor burns out. Since both V and R are unchanged for each resistor by the burned out resistor, I is also unchanged for each resistor, and the current in each resistor is still 12 A. Note that what does change is the total current on the circuit, from 48 A with 4 resistors (4 × 12 A) to 36 A with 3 resistors (3 × 12 A). Since the resistors are in parallel, fewer resistors will mean a higher equivalent resistance, so the total current is reduced, even though the current on each resistor is unchanged.

I don't understand how the total current could change without any change to a single resistor's current. Can someone please clarify. Thanks very much in advance.
 
Physics news on Phys.org


Let total resistance of parallel circuit is Rt.

With 4 identical resistors of resistance RΩ,
1/Rt =1/R1+ 1/R2+ 1/R3+ 1/R4
Rt=R/4

With 3 identical resistors of resistance RΩ,
Rt=R/3
 


Durfys said:
I don't understand how the total current could change without any change to a single resistor's current. Can someone please clarify. Thanks very much in advance.

Hi, welcome to physics forums Durfys! Normally on this forum, people need to show their attempt at figuring out the problem first, and write that down, so that we know how to help. But this is your first post, so maybe you didn't know that. No worries.

I'm not certain what you're stuck on. Is it this sentence from the book: Since the resistors are in parallel, fewer resistors will mean a higher equivalent resistance, so the total current is reduced, even though the current on each resistor is unchanged. Um, It is true that the current on each resistor is unchanged, but one of the resistors has been removed, due to it being burned out! So the circuit does change. Maybe the sentence in the book made it seem like the circuit did not change at all. But the circuit does change.
 


Hey, sorry about that. Thank you very much for the replies, after spending more time thinking about the question I now understand what I missed.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top