In analyzing a series circuit, it becomes necessary to find voltage drop across one or more of the resistances. A simple voltage drop relationship may be obtained by referring to the following figure.

The total current is given by,

$I=\frac{E}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}$

And the voltage drop are given by,

${{V}_{1}}=I{{R}_{1}}=E\frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}~~~~\text{ }~~~\left( 1 \right)$

${{V}_{2}}=I{{R}_{2}}=E\frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}~~\text{ }~~~~~~\left( 2 \right)$

${{V}_{3}}=I{{R}_{3}}=E\frac{{{R}_{3}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}~~\text{ }~~~~~~\left( 3 \right)$

Notice in these relationships that a voltage drop across a resistor is proportional to the ratio of that resistance to the total resistance of that circuit. A general statement called the voltage divider rule can be made as follows:

In a series circuit, the voltage drop across a particular resistor R_{n} is the source voltage times the fraction R_{n}/R_{t}.

${{V}_{n}}=E\frac{{{R}_{n}}}{{{R}_{t}}}$

In the above expression, E expresses source voltage and R_{t }is the total circuit resistance.

**Example of Voltage Division**

Determine the voltage drops V_{1}, V_{2}, and V_{3} in the circuit of the following Fig.

**Solution**

The total resistance

${{R}_{t}}=60+85+55=200~\Omega $

Then, by equations 1, 2 and 3

${{V}_{1}}=E\frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}=10*\frac{60}{200}=3~V$

${{V}_{2}}=E\frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}=10*\frac{85}{200}=4.25~V$

${{V}_{3}}=E\frac{{{R}_{3}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}=10*\frac{55}{200}=2.75~V$

A check by Kirchhoff’s voltage law indicates that

${{V}_{1}}+{{V}_{2}}+{{V}_{3}}=10~V=E$

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**Current Divider Circuit**

Resistors in parallel lend itself to a current division rule. In the following figure, the total current coming to the parallel combination of R_{1} and R_{2} divides into the currents I_{1} and I_{2 }respectively.

The branch currents are:

${{I}_{1}}=\frac{V}{{{R}_{1}}}$

${{I}_{2}}=\frac{V}{{{R}_{2}}}$

However, the voltage drop V equals I_{t}R_{t} where, R_{t} is given by the following equation

${{R}_{t}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}$

So, the voltage drop is,

$V={{I}_{t}}{{R}_{t}}={{I}_{t}}\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}$

Substitute V into branch currents, we have

${{I}_{1}}={{I}_{t}}\frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}~~~~~~~\text{ }\cdots \text{ }~~~~\left( 4 \right)$

${{I}_{2}}={{I}_{t}}\frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}~~~~~~~~\text{ }\cdots \text{ }~~~~\left( 5 \right)$

Equations (4) and (5) are mathematical statements for current divider rule, which is stated as follows:

“With two resistors in parallel, the current in either resistor is the total current times the ratio of the opposite resistor over the sum of the two resistors.”

**Example of current division**

Find the branch currents I_{1} and I_{2} for the parallel circuit of the following Fig**.**

**Solution**

By the current division principle,

${{I}_{1}}={{I}_{t}}\frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=20mA*\frac{5K}{5K+10K}=6.7~mA$

${{I}_{2}}={{I}_{t}}\frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}=20mA*\frac{10K}{5K+10K}=13.3~mA$

In above example, notice that since R_{1} is twice the resistance of R_{2}, one-half as much current will flow in it as in R_{2}. Thus is any parallel circuit, the ratio between any two branch currents equals the inverse of their resistance ratio.

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