Source Transformation
It is use to make the circuit simply. It transforms the source to make or having series or parallel in resistor to combine. In voltage, it transforms if the resistor connected in series while current transform if the resistor is parallel.
Source transformation also applied in dependent source, the process is still the same in independent source.
I learned that this method help us to make easier our circuit by transform it into a simple circuit which the source involved that can’t affect the value of circuit.
Superposition
Use to determine the specific value of contribution of each independent source. This is use in a circuit that has two or more independent source. For finding the contribution of a source independent first left one source and kill the rest independent source. For killing the voltage source make it short whiles in current make it open when it kills.
Find Vo distributed of each source and sum up all Vo to have the total Vo.
Kill the 2A source:
We can use Voltage divider for getting the Vo giving by 5V source.
Vo = Vo(10Ω)/5Ω +10Ω
Vo = 10/3 or 3.333V
Kill source 2A:
We can use current divider and ohm’s law.
I = 2(5)/10+5
I = 2/3 or 0.667A
Vo = IR
Vo = 0.667(10Ω)
Vo = 6.667V
In Real Life:
Superposition is just like the source where we get our supply like (Socoteco II, Meralco, Napocor, Davao Lights, Etc.) if one is shutdown the supply will decrease and consumption wills still the same. What will happen is having a short in supply. And the best solution is to less the supply in order to support long last. Or else it will vanish. It just like in General Santos City, we have a rotating brownout cause of some supplier shutdown or having maintenance.
thevenin’s theorem
this theorem state that the voltage in an open circuit is equivalent to Vth which is Vth=i(Rth) where Rth is the total resistance in the circuit that the source is killed.
Solve Rth :
Vth= Voltage in 3Ω
we can use voltage because 3Ω is connected in series with the source and parallel in a-b which is Vth :
Vth=5v(3Ω)/6Ω+3Ω
Vth=5/3 or 1.667V
Maximum Power Transfer
I conclude that Rth (thevenin’s resistance) is equal to the RL (load resistance) by using derivatives of the equation of power deliver to the RL (load resistance) .