# What is the voltage divider? (with examples)

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

*V _{i} = IZ _{i} .*

Where V _{i} is the voltage in the circuit element, I is the current flowing through it and Z _{i} corresponding impedance.

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

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

As the source represents a voltage increase, V = +12 V. And at 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:

+12V – 6V – 6V = 0V

That’s where the name of the voltage divider comes from, because, by means of series resistors, it is possible to easily obtain lower voltages from a source with a higher voltage.

__The voltage divider equation__

__The voltage divider equation__

Let’s continue to consider a purely resistive circuit. We know that the current I flowing through a series resistor circuit 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} … _{Ri} represents each series resistance of the circuit. Therefore:

V = I ∑ R _{i}

So the current ends up being:

I = V / ∑ R _{i}

Now let’s calculate the voltage on one of the resistances, the resistance R _{i,} for example:

V _{i} = (V / ∑ R _{i} ) R _{i}

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

**Voltage divider with 2 resistors**

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

And, in the special case where R _{1} = R _{2} , V _{I} = V / 2, regardless of current, as indicated at the beginning. This is the simplest voltage divider ever.

The following figure shows the layout of this divider, wherein V, the input voltage is symbolized as V _{in} and V _{i} is the voltage obtained by dividing the voltage between resistors R _{1} and R _{2} .

__Solved Examples__

__Solved Examples__

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

**– Example 1**

A 12 V source is available, which has to be split into 7 V and 5 V, via two resistors R _{1} and R _{2} . There is a fixed resistance of 100 Ω and a variable resistance whose range is between 0 and 1kΩ. What are the options for configuring the circuit and setting the value of resistance R _{2} ?

**Solution**

To solve this exercise, the voltage divider rule for two resistors will be used:

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

Unknown resistance R _{2} must be at 5 V:

YR _{1} to 7 V:

5 (R _{2} 100) 12 = R _{2}

500 = 7 R _{2}

R _{2} = 71.43 Ω

Likewise, the other equation can be used to get the same value, or the result obtained can be substituted to check equality.

If now the fixed resistance is set to R _{2} , then R _{1 will} be at 7 V:

5 (100 + R _{1} ) = 100 x 12

500 + 5R _{1} = 1200

R _{1} = 140 Ω

Likewise, it is possible to check whether this value meets the second equation. Both values are in the variable resistance range, therefore, it is possible to implement the requested circuit in both modes.

**– example 2**

A DC direct current voltmeter for measuring voltages within a certain range is based on the voltage divider. To build this voltmeter, you need a galvanometer, for example, the one from D’Arsonval.

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

The galvanometer has an internal resistance R _{L} maximum current which tolerates only a small current, called I _{L} . Therefore, the voltage across the galvanometer is V _{m} = I _{L} R _{L} .

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 gauge, the voltage to be measured must not exceed the maximum allowable, which is the maximum needle deflection that the device has. But we assume that V _{m} is small because I _{G } and R _{G } are small.

However, when the galvanometer is connected in series with another resistor R _{S} , called the *limiting resistance* , we can increase the measuring range of the galvanometer from small V _{m} to a certain higher voltage ε. When this tension is reached, the instrument needle undergoes maximum deflection.

The design scheme is as follows:

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

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

**1V full scale voltmeter**

For example, suppose the internal resistance of the galvanometer is R _{G} = 50 Ω and the maximum current it can withstand is I _{G} = 1 mA, the limiting resistance RS of the voltmeter built with this galvanometer to measure a maximum voltage of 1 V is calculated. Thus:

I _{G} (R _{S} + R _{L} ) = 1 V

R _{S} = (1 V / 1 x 10 ^{-3} A) – R _{G}

R _{S} = 1000 Ω – 50 Ω = 950 Ω