I'm having trouble trying to understand voltage. From my understanding, it is the amount of energy required to move one Coulomb of charge from one point to another (J/C). So if you have a voltage of 9v across a resistor, then you are saying that it takes 9 joules of energy to move one Coulomb of charge through that resister. If the voltage is 10v, it takes 10 joules to move the one coulomb and so on
But this raises two questions for me with the above circuit
1) why is it that no matter what the value of the resistor is, it always takes 9 joules of energy to move one coulomb of charge through it? If the resistor value is increased, then surely it would take more energy to move the coulomb of charge through it? (I know this is ohms law, but I want to understand what the electorons are doing)
2) why is it that all the voltage is dropped before reaching the end of a loop (KVL) ? or rather, why do the electrons always lose all their energy before coming backing around to the positive terminal? howcome they don't retain some of their energy when there is little resistance?
I've tried to answer the first myself :
1) As the voltage is fixed (meaning that each electron has the same potential energy no matter how complicated or how much resistance the circuit has), a higher resistance does not affect the amount of energy it takes to move an electron through the resistor, but rather, how long it takes to move through the resistor (I = V/R).
A higher resistance means a lower current, in that fewer electrons are passing through the resistor per second, thus a lower C/S rate
A lower resistance means a higher current, in that more electrons are passing through the resistor per second, thus a higher C/S rate
In both cases - the same amount of work is done to move one coulomb. Just over a different time range.
By increasing the resistance, you're reducing the current, thus decreasing the amount of energy that can be consumed from the battery per second (P = IV)
But this raises two questions for me with the above circuit
1) why is it that no matter what the value of the resistor is, it always takes 9 joules of energy to move one coulomb of charge through it? If the resistor value is increased, then surely it would take more energy to move the coulomb of charge through it? (I know this is ohms law, but I want to understand what the electorons are doing)
2) why is it that all the voltage is dropped before reaching the end of a loop (KVL) ? or rather, why do the electrons always lose all their energy before coming backing around to the positive terminal? howcome they don't retain some of their energy when there is little resistance?
I've tried to answer the first myself :
1) As the voltage is fixed (meaning that each electron has the same potential energy no matter how complicated or how much resistance the circuit has), a higher resistance does not affect the amount of energy it takes to move an electron through the resistor, but rather, how long it takes to move through the resistor (I = V/R).
A higher resistance means a lower current, in that fewer electrons are passing through the resistor per second, thus a lower C/S rate
A lower resistance means a higher current, in that more electrons are passing through the resistor per second, thus a higher C/S rate
In both cases - the same amount of work is done to move one coulomb. Just over a different time range.
By increasing the resistance, you're reducing the current, thus decreasing the amount of energy that can be consumed from the battery per second (P = IV)
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