In part 4, we learned from Ohm’s law that resistance reduces the amount of current that flows through a circuit, but it turns out, voltage is also reduced.
In fact, Kirchhoff’s voltage law states that the sum of all voltage drops around a circuit loop is equal to the sum of all voltage sources. In other words, if we had a circuit loop as in following illustration, with a
5V voltage source, the total drop in voltage across all the resistors would be
Algebraically, we can express this as:
Vs = V1 + V2 + V3
V1 represents the voltage drop at
R1, and so forth.
This means that at point
A, relative to ground, the voltage would be
5V, and at point
D, the voltage is
OV. It turns out, that the voltage through each resistor is proportionate to that resistor’s part of the overall resistance. So if that resistance represents one half of the total resistance, then the voltage drop through that resistor will also be half of the total voltage drop, or
5V * .5 = 2.5V.
Deriving from Ohm’s Law
We can calculate Kirchhoff’s voltage law from Ohm’s law.
Recall that in a series resistance, each resistor experiences the same amount of current:
In this case, the total resistance (series resistances are additive), is
20Ω, therefore the current is:
Given: I = V / R Therefore: I = 5V / 20Ω = 0.25A = 250mA
So each resistor sees
0.25A of current.
Now, recalling Ohm’s law solved for voltage (
V = I * R), we can calculate the voltage drop for that current at each resistor:
R1 Voltage = 0.25A * 2Ω = 0.5V R2 Voltage = 0.25A * 6Ω = 1.5V R3 Voltage = 0.25A * 12Ω = 3V
If we add each of these voltage drops up:
Total voltage drops = 0.5V + 1.5V + 3V = 5V = Voltage Source
The total amount of voltage drop is the exact same amount as the voltage source!
Revisiting the previous circuit, if we put a voltmeter between points
D, and ground, we’d get
This is because the voltage drop is removed from the source voltage to get the output voltage after each resistor:
B -> Common/Ground = 5V - 3V = 2V C -> Common = 2V - 1.5V = .5V D -> Common = .5V - .5V = 0V
Voltage drop is often referred to as forward voltage (Vf). In many components, their voltage drop is actually the same amount of voltage that they require to work. For example, lights such as LEDs require a particular forward voltage to be met in order for them to have enough energy to work.
Simplifying Series Resistance
Since we know that resistors in series are additive, if we want to know the voltage drop at any point between resistors in series (as in the above diagram), we can simplify the previous circuit by considering any series resistances as a single resistor by adding them up, as illustrated in the following circuit diagram:
Note that the sigma (
∑) symbol means mathematical sum, and
Simplifying circuits like this, by combining resistances, is a common way to analyze circuits because it makes them more practical to deal with. In fact, when dealing with circuits, we often simplify them into single values that describe their overall resistance, and capacitance, which we’ll explore later. By simplifying circuits, we can more effectively design complex systems that incorporate multiple circuits into a larger design.
Using the same numbers from before, calculate the forward voltage (voltage drop) at
Given: Vf = I * R1 Therefore: Vf = 0.25A * 18Ω = 4.5V
That means that there is only
0.5V from the original
5V of electromotive force at point
B, since we have to remove the drop from the source voltage:
V (at point B) = 5V - 4.5 = 0.5V
Voltage Drop (Vf) of R2 is Vout
0.5V is ALSO the voltage drop from
C. This becomes a little clearer if we simplify the circuit diagram even further:
In this case, we see that
Vout = VS - R1Vf and
R2Vf = VS - R1Vf, which means that
Vout = R2Vf. Therefore, if we want to calculate
Vout, we can simplify things by calculating the voltage drop of
Voltage Drop is Proportionate
Because Ohm’s law is proportionate, the ratio of resistances determines how much the voltage is reduced at each interval. We can prove this by starting with Ohm’s law, solved for
Given: I = V / R Therefore I = Vs / (R1 + R2)
R2Vf, we solve for the voltage drop (
R2 and substitute
Vs / (R1 + R2) in for
Given: R2Vf = I * R2 And: I = Vs / (R1 + R2) Therefore: Vout = R2VF = (Vs / (R1 + R2)) * R2
We can test this using the values from before:
R2Vf = (5v / (18Ω + 2Ω)) * 2Ω = 0.5V
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