This law is also known as Point Law or Current law. In any electrical network , the algebraic sum of incoming currents to a point and outgoing currents from that point is Zero. Or the entering currents to a point are equal to the leaving currents of that point. In other words, the sum of the currents flowing towards a point is equal to the sum of those flowing away from it.
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By loop, I mean any path traced from one point in a circuit around to other points in that circuit, and finally back to the initial point. If we were to take that same voltmeter and measure the voltage drop across each resistor , stepping around the circuit in a clockwise direction with the red test lead of our meter on the point ahead and the black test lead on the point behind, we would obtain the following readings: We should already be familiar with the general principle for series circuits stating that individual voltage drops add up to the total applied voltage, but measuring voltage drops in this manner and paying attention to the polarity mathematical sign of the readings reveals another facet of this principle: that the voltages measured as such all add up to zero: In the above example, the loop was formed by the following points in this order: This is because the resistors are resisting the flow of electric charge being pushed by the battery.
Here we see what a digital voltmeter would indicate across each component in this circuit, black lead on the left and red lead on the right, as laid out in horizontal fashion: If we were to take that same voltmeter and read voltage across combinations of components, starting with the only R1 on the left and progressing across the whole string of components, we will see how the voltages add algebraically to zero : The fact that series voltages add up should be no mystery, but we notice that the polarity of these voltages makes a lot of difference in how the figures add.
That we should end up with exactly 0 volts across the whole string should be no mystery, either. Looking at the circuit, we can see that the far left of the string left side of R1: point number 2 is directly connected to the far right of the string right side of battery: point number 2 , as necessary to complete the circuit.
Since these two points are directly connected, they are electrically common to each other. And, as such, the voltage between those two electrically common points must be zero. Note how it works for this parallel circuit: Being a parallel circuit, the voltage across every resistor is the same as the supply voltage: 6 volts. Tallying up voltages around loop , we get: Note how I label the final sum voltage as E Since we began our loop-stepping sequence at point 2 and ended at point 2, the algebraic sum of those voltages will be the same as the voltage measured between the same point E , which of course must be zero.
All we have to do to comply with KVL is to begin and end at the same point in the circuit, tallying voltage drops and polarities as we go between the next and the last point.
The two series circuits share a common wire between them wire , making voltage measurements between the two circuits possible.
If we wanted to determine the voltage between points 4 and 3, we could set up a KVL equation with the voltage between those points as the unknown:.
Kirchhoff’s Voltage Law (KVL)
Example 1: Find the three unknown currents and three unknown voltages in the circuit below: Note: The direction of a current and the polarity of a voltage can be assumed arbitrarily. To determine the actual direction and polarity, the sign of the values also should be considered. For example, a current labeled in left-to-right direction with a negative value is actually flowing right-to-left. All voltages and currents in the circuit can be found by either of the following two methods, based on KVL or KCL respectively. The loop-current method mesh current analysis based on KVL: For each of the independent loops in the circuit, define a loop current around the loop in clockwise or counter clockwise direction.
Kirchhoff’s Current & Voltage Law (KCL & KVL) | Solved Example