Voltage Dividers
concept
When you design a circuit a VERY big part of it is creating the right voltage at the right point. You can't just have 100 batteries of different voltages sitting by in order to set the right voltage for each point in the circuit. The voltage divider is a basic building block of probably every circuit. The idea behind the voltage divider is that you can take a battery and "send" some of the voltage off to one part of the circuit and "send" the rest to another part.
Alternatively you can use a battery larger than you need and, with a voltage divider, only have the voltage you actually want interacting with the rest of your circuit.
It might not seem like a big deal now, but the voltage divider is a very large part of every piece of technology in the world.
fact
The basic voltage divider consists of a battery connected to two series resistors. The voltage between the two resistors is some positive amount that is lower than the battery voltage. You control the fraction of the battery voltage that appears between the resistors by setting the ratio of the two resistors' resistances.
The basic voltage divider circuit looks like this:
fact
The voltage \(V_o\) is given by the equation:
\(V_o = V \frac{R_2}{R_1 + R_2}\)
derivation
Here we'll prove that the voltage divider works. You don't need to know this but it'll help you work problems later.
We start with the circuit below:
Here \(V_o\) represents the voltage between \(R_1\) and \(R_2\). We draw it that way because in a real circuit we'd have more "stuff" connected to that open circle under \(V_o\) that would use the \(V_o\) voltage we just created.
Our first step will be to find the current \(I\) running through this series circuit.
A simple application of series resistors and Ohm's Law will show that \(I = \frac{V}{R_1 + R_2}\)
Now to find the voltage at \(V_o\) we're going to notice that the voltage at the bottom of \(R_2\) MUST be 0V since we haven't drawn in a ground node and thus the back of the battery is the ground node, which is the same point in the circuit as the bottom of \(R_2\) (if that confuses you just remember that wires in a circuit diagram are just to space things out, everything connected to the same wire is the same point in the circuit)
So if the bottom of \(R_2\) is 0V and we know the current through it then the top of \(R_2\) must have a voltage given by:
\(V_o = I R_2 = \frac{V}{R_1 + R_2} R_2 = V \frac{R_2}{R_1 + R_2}\)
The process we just went through is used in analysing lots of different circuits. Try to understand each step and why we did it rather than memorising each of the moves. Your ability to learn electronics later on will be greatly improved.
Here \(V_o\) represents the voltage between \(R_1\) and \(R_2\). We draw it that way because in a real circuit we'd have more "stuff" connected to that open circle under \(V_o\) that would use the \(V_o\) voltage we just created.
Our first step will be to find the current \(I\) running through this series circuit.
A simple application of series resistors and Ohm's Law will show that \(I = \frac{V}{R_1 + R_2}\)
Now to find the voltage at \(V_o\) we're going to notice that the voltage at the bottom of \(R_2\) MUST be 0V since we haven't drawn in a ground node and thus the back of the battery is the ground node, which is the same point in the circuit as the bottom of \(R_2\) (if that confuses you just remember that wires in a circuit diagram are just to space things out, everything connected to the same wire is the same point in the circuit)
So if the bottom of \(R_2\) is 0V and we know the current through it then the top of \(R_2\) must have a voltage given by:
\(V_o = I R_2 = \frac{V}{R_1 + R_2} R_2 = V \frac{R_2}{R_1 + R_2}\)
The process we just went through is used in analysing lots of different circuits. Try to understand each step and why we did it rather than memorising each of the moves. Your ability to learn electronics later on will be greatly improved.
example
Find the voltage \(V_o\) in the circuit below
practice problems