- Current Divider Rule Definition: The current divider rule calculates the current through each parallel path in a circuit, based on the impedances of each path.
- Voltage Divider Formula: The voltage across any impedance in a series circuit can be found by multiplying the total voltage by the ratio of the target impedance to the total impedance.
- Practical Application: Example problems show how to apply the current and voltage divider rules to find specific currents and voltages in circuit diagrams.
- Electrical Impedance: Understanding impedance is crucial, as it influences how current and voltage are divided in circuits.
- Kirchhoff’s Current Law: This law helps to establish the balance of current at a junction, supporting the calculations for current division.
Current Division Rule
Parallel branches have the same voltage. Their currents therefore depend on their impedances: a lower impedance takes a larger current magnitude for a given branch voltage.
For several parallel branches, write each branch admittance as Y = 1/Z. The current in branch k is the total current multiplied by Yk divided by the sum of all branch admittances.
The two-branch form of this relationship is the current divider rule. It provides a quick circuit-analysis tool in electric engineering when the parallel branch impedances and total current are known.
For sinusoidal AC analysis, use complex impedances and current phasors at the same frequency. The branch current can differ from the total current in both magnitude and phase.
Consider two impedances Z1 and Z2 connected in parallel.
The total current I divides into I1 and I2. Currents I1 and I2 pass through Z1 and Z2, respectively. The goal is to express I1 and I2 in terms of I, Z1, and Z2.
Because Z1 and Z2 are parallel, they have the same voltage drop. Their voltage-current relations give
Applying Kirchoff’s current law at the junction gives the total-current relation below.
The two equations determine I1 and I2.
From equation (1),
Substitution in equation (2) gives 
Rearranging,
then,
and,
Therefore,
Substituting I1 into the current sum gives 
Thus, I1 and I2 are expressed in terms of I, Z1, and Z2. Each two-branch formula has the other branch’s impedance in the numerator.
The following steps apply the two-branch rule.
To determine I1, identify the total current and both branch impedances.
Substitution in the current-divider expression gives the next result.
The next example applies the rule to two parallel impedances.
Applying the current division rule gives:
Here I1 is the current through Z1.
Substituting the stated values gives:
For the second branch:
I2 can also be found by subtracting I1 from the total current:
The branch currents should add vectorially to the source current. For DC resistors they have the same phase, so ordinary signed addition is sufficient.
Voltage Division Rule
The voltage division rule finds the voltage across one impedance in a series chain. Suppose Z1, Z2, Z3,…..Zn are in series and a voltage source V is connected across the complete chain.
The voltage divider circuit is shown below:
To find the voltage across Z3, first add the series impedances Z1, Z2, Z3 …. Zn. The effective impedance Zeff seen by the source is
The series current is
The same current passes through every series impedance, so the voltage across Z3 is
The voltage across Z1 is likewise
For any element k,
Here k = 1, 2, 3,….n, and Z1, Z2, Z3 ,…….Zn must form the same series path. Any load connected across one element changes that branch’s effective impedance and must be included before applying the rule.
The voltage division rule therefore multiplies the source voltage by the selected impedance divided by the total series impedance. With complex AC impedances, this ratio can change both magnitude and phase.
The general voltage-divider expression is
The following example uses the voltage division rule to find the voltage across each impedance.
Voltage Division Rule Example Problem
These impedances are connected in series:
A 100 V source is connected across the complete series chain. Find the voltage across each impedance.
Solution:
Applying the voltage-divider expression gives
For the next element,
Vz3 can also be found from Kirchhoff’s voltage law after the other two voltages are known.
This subtraction method must preserve voltage polarity and, for AC circuits, phasor phase.
Now consider equal series impedances:
The voltage across each equal impedance is:
When n equal impedances are in series and no unmodelled load changes a branch, each receives V/n of the source voltage. Equal magnitude alone is insufficient if the complex impedance angles differ.
