Superposition Theorem is fairly easy to understand. Given the following example:
Superposition applies when more than one source existed in the circuit and in this circuit, there are two voltage sources and one current source and therefore this theorem can be applied. Let’s assume that we need to find the current in the circled element due to the three sources and we analyse the effect of each of the sources and then combine (algebraic sum) them. This is what the superposition theorem’s approach is.
When we focus on only one source at a time, the other voltage source will be brought to 0 volt which means a short circuit with 0 ohm resistance (ohm’s law says V=IR, V=0 when R=0), on the other hand, the other current source will be brought to 0 ampere which means an open circuit as open circuit does not allow any current, ie 0.
Therefore, when we analyse the circuit due to voltage source V1, the circuit will become:
To find the current in R2, we need to find the total RT which is 1k(R1) in series with 1.5k(R2) in parallel with 3k(R3) as the voltage source V1=12V is experiencing this resistance. RT=1k+1.5k×3k/(1.5k+3k)=1k+1k=2k.
The current drawn from V1 of 12V is I=12V/2k=6mA. This current will be divided among R2(1.5k) and R3(3k) using current divider rule:
Similarly, finding IR2 due to V2(36V) using the same technique and find IR2 due to I1(5mA).
In this case, RT=3k+1.5k//1k=3.6k and then IT=-36V/3.6k=-10mA
Using current divider, IR2=-10mA×G2/GT=10mA×(1/1.5k)/(1/1.5k+1/1k)=-4mA.
In the case of 5mA current source, all resistors are in parallel, current divider rule will solve the problem. IR2=5mA×(1/1.5k)/(1/1.5k+1/1k+1/3k)=1.7mA
Combining all the IR2 due to the 3 sources, you will get the overall IR2 of the circuit using this superposition theorem.
The solution in here is then the algebraic sum: 1.7mA-4mA+4mA=1.7mA