This question comes from berkley review E&M. Picture attached below.
It asks: if you add ad resistor in parallel, then what happens to the current through the pre-existing resistors?
The answer is that the current remains unchanged.
From one viewpoint, I understand, because no matter how many resistors you add to a circuit, the voltage across each resistor will remain the same (crazy right?), which means that the current, since V = IR , will be the same as well, even if another resistor is added.
But what I dont understand is, the junction rule also tells us that current going into a split and going out of a split must be equal, so intuitively, if you add another lane for current to flow through, then the current through the other lanes should decrease.
However we know this is not true, because in adding another resistor in parallel ,the equivalent resistance in the circuit decreases, so the current going into the circuit goes up, just enough to offset the new current flowing into the added resistor.
My issue here is that, when i considered this problem, i first though of the how current going into a split must be equal to current going out, and by that logic, the current through the pre-existing resistors should decrease. Why doesn't that logic apply?
It asks: if you add ad resistor in parallel, then what happens to the current through the pre-existing resistors?
The answer is that the current remains unchanged.
From one viewpoint, I understand, because no matter how many resistors you add to a circuit, the voltage across each resistor will remain the same (crazy right?), which means that the current, since V = IR , will be the same as well, even if another resistor is added.
But what I dont understand is, the junction rule also tells us that current going into a split and going out of a split must be equal, so intuitively, if you add another lane for current to flow through, then the current through the other lanes should decrease.
However we know this is not true, because in adding another resistor in parallel ,the equivalent resistance in the circuit decreases, so the current going into the circuit goes up, just enough to offset the new current flowing into the added resistor.
My issue here is that, when i considered this problem, i first though of the how current going into a split must be equal to current going out, and by that logic, the current through the pre-existing resistors should decrease. Why doesn't that logic apply?