Kirchhoff’s voltage law when officially stated is the algebraic sum of the potential differences in a circuit loop must be zero. Potential rises are passive while potential drafts are negative. It can algebraically be described in the formula below. The law sounds more complicated than it actually is. Generally speaking it says around any loop in a circuit the voltage rise must equal the voltage drops. Another way of thinking about this is to consider that whatever energy a charge starts with in a circuit loop it must end up losing all the energy by the time it gets to the end, or we could say by the time a charge makes it to the end of a circuit it must have given all its energy to to do work.
Let’s look at some diagrams that represent several possible circuits or loops within a circuit. This is a sample circuit showing the potential differences across the source and the resister. According to Kirchhoff’s law the sum of the potential differences will be zero. Now let’s check to see if the voltage measurements equal zero. Solve for the voltage drop, given the power source has the voltage of positive 50. That’s correct. The voltage drop needs to be negative 50 or the sum of the potential differences to equal zero. Sorry, that is incorrect. The voltage drop needs to be negative 50 or the sum of the potential differences to equal zero.
This diagram shows the potential differences from one side of the resistor to the other. Notice the resistance is actually the difference between one potential to another.
This animation shows the same circuit but only looks at the potential differences as you go around the loop. Again Kirchhoff’s voltage law says the sum of the potential differences has to be zero. Now let’s check to see if the voltage measurements equal zero. Solve for the voltage drop where the second resistor given the power source has a voltage of plus 100 and the first voltage drop has the value of negative 20. That’s correct. The voltage drop needs to be negative 80 or the sum of the potential differences to equal zero. This concludes this demonstration. Sorry, that’s incorrect. The voltage drop needs to be negative 80 or the sum of the potential differences to equal zero. This concludes this demonstration.
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