# Substitution Theorem

As name implies, the main concept of this theorem which is based upon substitution of one element by another equivalent element. ** Substitution theorem** gives us some special insights in circuit behavior. This theorem is also used to prove several other theorems.

## Statement

**Substitution theorem**states that if an element in a network is replaced by a voltage source whose voltage at any instant of time is equals to the voltage across the element in the previous network then the initial condition in the rest of the network will be unaltered. Or alternately if an element in a network is replaced by a current source whose current at any instant of time is equal to the current through the element in the previous network then the initial condition in the rest of the network will be unaltered.

## Explanation

Let us take a circuit as shown in fig - a, Let, V is supplied voltage and Z_{1}, Z

_{2}and Z

_{3}is different circuit impedances. V

_{1}, V

_{2}and V

_{3}are the voltages across Z

_{1}, Z

_{2}and Z

_{3}impedance respectively and I is the supplied current whose I

_{1}part is flowing through the Z

_{1}impedance whereas I

_{2}part is flowing through the Z

_{2}and Z

_{3}impedance.

Now if we replace Z_{3} impedance with V_{3} voltage source as shown in fig-b or with I_{2} current source as shown in fig-c then according to **Substitution Theorem** all initial condition through other impedances and source will remain unchanged.
i.e. - current through source will be I, voltage across Z_{1} impedance will be V_{1}, current through Z_{2} will be I_{2} etc.

### Example of Substitution Theorem

For more efficient and clear understanding let us go through a simple practical example: Let us take a circuit as shown in fig - dAs per voltage division rule voltage across 3Ω and 2Ω resistance are
If we replace the 3Ω resistance with a voltage source of 6 V as shown in fig - e, then
According to Ohm’s law the voltage across 2Ω resistance and current through the circuit is
Alternately if we replace 3Ω resistance with a current source of 2A as shown in fig - f, then
Voltage across 2Ω is V_{2Ω} = 10 - 3× 2 = 4 V and voltage across 2A current source is V_{2A} = 10 - 4 = 6 V
We can see the voltage across 2Ω resistance and current through the circuit is unaltered i.e all initial condition of the circuit is intact.