# Electric Current and Voltage Division Rule

## Current Division Rule

When current flows through more than one parallel paths, each of the paths shares a definite porion of the total current depending upon the impedance of that path. The definite portion of total current shared by any of the parallel paths can easily be calculated if the impedance of that path and the equivalent impedance of the parallel system are known to us. The rule or formula derived from these known impedances to know the portion of total current through any parallel path is known as**current division rule**. This rule is very important and widely used in the field of electric engineering in different applications.

Actually this rule finds application when we have to find the current passing through each impedance when these are connected in parallel. Let us say, two impedances Z_{1} and Z_{2} are connected in parallel as shown below.
A current I passes and is being divided into I_{1} and I_{2} at the junction of these two impedances as shown. I_{1} and I_{2} pass through Z_{1} and Z_{2} respectively. Our aim is to determine I_{1} and I_{2} in terms of I, Z_{1} and Z_{2}.
As Z_{1} and Z_{2} are connected in parallel, voltage drop across each will be same. Hence, we can write
Also applying Kirchoff’s current law at junction, we get
We have two equations and can determine I_{1} and I_{2}.

From (1), we have
Putting this in (2), we get
or,
or,
or,
We have
Putting value of I_{1}, we get
Thus, we have determined I_{1} and I_{2} in terms of I, Z_{1} and Z_{2}.

This rule is applied as follows.

Suppose we have to determine I_{1}. We proceed as
Applying above rule, we will get

Applying

**current divison rule**, we will have Where, I

_{1}= current passing through Z

_{1}.

Putting given numerical values, we get Similarly, The other way to find I

_{2}is as This is how we can apply current division rule.

## Voltage Division Rule

**Voltage division rule**is applied when we have to find voltage across some impedance. Let us assume that the impedances Z

_{1}, Z

_{2}, Z

_{3},…..Z

_{n}are connected in series and voltage source V is connected across them as shown below. Our aim is to find voltage across some impedance, say, Z

_{3}. We see that Z

_{1}, Z

_{2}, Z

_{3}…. Z

_{n}are connected in series. Hence, effective impedance Z

_{eff}as seen by the voltage is given by Current passing the circuit is given by This current is passing through all the impedances connected in series. Hence, voltage across Z

_{3}is given by Similarly, voltage across Z

_{1}will be given by In general, we can write Where, k = 1, 2, 3,….n and impedances Z

_{1}, Z

_{2}, Z

_{3},…….Z

_{n}should be connected in series.

This is called

**voltage division rule**and frequently used to determine the voltage across some impedance. We can write this rule in words as given below.

Voltage across some impedance

**voltage divisio rule**.

Problem

impedance are connected in series. Across these impedance connected in series, a voltage source of 100V is connected as shown below. Determine the voltage across each impedance. Solution:

Applying voltage division rule, we get Similarly, We can also determine V

_{z3}as follows. Actually, we can determine voltage across any impedance in this way if voltages across all other remaining impedances are known.

When we , voltage across each impedance is given by Thus voltage will be same across each impedance and it equals V/n, that is, source voltage divided number of impedances connected in series.

**Comments/Feedbacks**

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Sathya Sunduposted this comment on 31-05-17 09:26:57 amIn current division example value of I

_{1}is wrong.I

_{1}= 6.49 - 0.5j