# What Is The Voltage Divider? (with Examples)

The voltage divider or voltage divider consists of an association of resistors or impedances in series connected to a source. In this way, the voltage V supplied by the source -input voltage- is distributed proportionally in each element, according to Ohm’s law:

V i = IZ i .

Where V i is the voltage across the circuit element, I is the current flowing through it and Z i the corresponding impedance.

When arranging the source and the elements in a closed circuit, Kirchhoff’s second law must be fulfilled, which states that the sum of all the voltage drops and rises is equal to 0.

For example, if the circuit to be considered is purely resistive and a 12 volt source is available, simply by placing two identical resistors in series with said source, the voltage will be divided: each resistance will have 6 Volts. And with three identical resistors you get 4 V in each one.

Since the source represents a voltage rise, then V = +12 V. And in each resistor there are voltage drops that are represented by negative signs: – 6 V and – 6 V respectively. It is easy to see that Kirchoff’s second law is fulfilled:

+12 V – 6 V – 6 V = 0 V

This is where the name voltage divider comes from, because through series resistors, lower voltages can easily be obtained starting from a source with a higher voltage.

## The voltage divider equation

Let’s continue considering a purely resistive circuit. We know that the current I through a circuit of series resistors connected to a source as shown in figure 1 is the same. And according to Ohm’s law and Kirchoff’s second law:

V = IR 1 + IR 2 + IR 3 +… IR i

Where R 1 , R 2 … R i represents each series resistance of the circuit. Thus:

V = I ∑ R i

So the current turns out to be:

I = V / ∑ R i

Now let’s calculate the voltage across one of the resistors, the resistor R i for example:

V i = (V / ∑ R i ) R i

The previous equation is rewritten as follows and we have the voltage divider rule ready for a battery and N resistors in series:

$V_{i}=\left&space;(\frac{R_{i}}{\sum_{i=1}^{N}R_{i}}&space;\right&space;)V$

### Voltage divider with 2 resistors

If we have a voltage divider circuit with 2 resistors, the above equation becomes:

# Original text

$V_{i}=\left&space;(\frac{R_{1}}{R_{1}+R_{2}}&space;\right&space;)V$

And in the special case where R 1 = R 2 , V i = V / 2, regardless of the current, just as it was said at the beginning. This is the simplest voltage divider of all.

In the following figure is the diagram of this divider, where V, the input voltage, is symbolized as V in , and V i is the voltage obtained by dividing the voltage between the resistors R 1 and R 2 .

## Worked Examples

The voltage divider rule will be applied in two resistive circuits to obtain lower voltages.

### – Example 1

A 12 V source is available, which has to be divided into 7 V and 5 V by two resistors R 1 and R 2 . There is a 100 Ω fixed resistance and a variable resistance whose range is between 0 and 1kΩ. What options are there to configure the circuit and set the value of the resistor R 2 ?

#### Solution

To solve this exercise we will use the rule of the voltage divider for two resistors:

$V_{i}=\left&space;(\frac{R_{1}}{R_{1}+R_{2}}&space;\right&space;)V$

Suppose that R 1 is the resistance found at a voltage of 7 V and there is placed the fixed resistance R 1 = 100 Ω

The unknown resistance R 2 must be at 5 V:

$5=\left&space;(\frac{R_{2}}{100+R_{2}}&space;\right&space;)12$

YR 1 to 7 V:

$7=\left&space;(\frac{100}{100+R_{2}}&space;\right&space;)12$

5 (R 2 +100) = 12 R 2

500 = 7 R 2

R 2 = 71.43 Ω

You can also use the other equation to obtain the same value, or substitute the result obtained to check for equality.

If now the fixed resistance is placed as R 2 , then R 1 is at 7 V:

$5=\left&space;(\frac{100}{100+R_{1}}&space;\right&space;)12$$7=\left&space;(\frac{R_{2}}{100+R_{2}}&space;\right&space;)12$

5 (100 + R 1 ) = 100 x 12

500 + 5R 1 = 1200

R 1 = 140 Ω

In the same way, it is possible to verify that this value satisfies the second equation. Both values ​​are in the range of the variable resistor, therefore it is possible to implement the requested circuit in both ways.

### – Example 2

A DC direct current voltmeter to measure voltages in a certain range, is based on the voltage divider. To build such a voltmeter, a galvanometer is required, for example D’Arsonval’s.

It is a meter that detects electrical currents, equipped with a graduated scale and an indicating needle. There are many models of galvanometers, the one in the figure is a very simple one, with two connection terminals that are on the back.

The galvanometer has an internal resistance R G maximum current, which tolerates only a small current, called I G . Consequently, the voltage across the galvanometer is V m = I G R G .

To measure any voltage, the voltmeter is placed in parallel with the element to be measured and its internal resistance must be large enough not to draw current from the circuit, otherwise it will alter it.

If we want to use the galvanometer as a meter, the voltage to be measured must not exceed the maximum allowed, which is the maximum deflection of the needle that the device has. But we assume that V m  is small, since I G   and R are.

However, when the galvanometer is connected in series with another resistor R S , called a limiting resistor , we can extend the measurement range of the galvanometer from the small V m to some higher voltage ε. When this voltage is reached, the instrument needle experiences maximum deflection.

The design scheme is as follows:

In figure 4 on the left, G is the galvanometer and R is any resistance over which you want to measure the voltage V x .

The figure on the right shows how the circuit with G, R G and R S is equivalent to a voltmeter, which is placed in parallel to the resistance R.

#### 1V Full Scale Voltmeter

For example, suppose that the internal resistance of the galvanometer is R G = 50 Ω and the maximum current that it supports is I G = 1 mA, the limiting resistance RS for the voltmeter built with this galvanometer to measure a maximum voltage of 1 V is calculated So:

I G (R S + R G ) = 1 V

R S = (1 V / 1 x 10 -3 A) – R G

R S = 1000 Ω – 50 Ω = 950 Ω

## References

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6. Hyperphysics. Design of a voltmeter. Recovered from: hyperphysics.phy-astr.gsu.edu.
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