What is the superposition theorem?
In a linear time-invariant circuit with multiple voltage and current sources, the superposition theorem conveniently obtains the node voltage and branch current.
The superposition theorem states, ” The total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately. “
To find the current from Rload, we may apply the superposition theorem. The method is more intuitive and less complex than using KCL and KVL.
![superposition_theorem-1 superposition_theorem-1](https://analogcircuitdesign.com/wp-content/uploads/elementor/thumbs/superposition_theorem-1-e1703358827914-qh8y0e8ghr81usw07smr8faretsfqu6p714k4h8hz4.png)
Keep only one power source "Turn on". "Turn off" all other power sources.
To make sure that only one power source is “on” at a time and the rest are turned off:
- We are replacing all other independent voltage sources with a short circuit (using a zero resistance wire).
- Replacing all other independent current sources with an open circuit (circuit break having infinite resistance).
![superposition_theorem_V1-1 superposition theorem. creating an open where current source was present](https://analogcircuitdesign.com/wp-content/uploads/elementor/thumbs/superposition_theorem_V1-1-e1703358865598-qh8y1dybpgky3zg4f82kv6a9zgwdvc4hzxx0czrjfi.png)
![superposition_theorem_I1-1 Superposition theorem. Creating shorts where voltage sources were present](https://analogcircuitdesign.com/wp-content/uploads/elementor/thumbs/superposition_theorem_I1-1-e1703358895242-qh8y265hehnjsab5uk9dxz63t11ea98g3thkralqbc.png)
![superposition_theorem_V2-1 superposition_theorem_V2-1](https://analogcircuitdesign.com/wp-content/uploads/elementor/thumbs/superposition_theorem_V2-1-e1703358925844-qh8y2ycn3iq5gl679wg70s1xml6ep6ce7p255lfx5i.png)
Calculate the current and voltage (with direction) because of all the sources
Current from Rload because of V1 from terminal A to terminal B (Fig. 2),
$$I_{V1}=\cfrac{V_1}{R_1+R_{load}||(R_2+R_4)}\cfrac{R_2+R_4}{R_{load}+R_2+R_4}$$
Current from Rload because of I1 from terminal A to terminal B (Fig. 3),
$$I_{I1}=-I_1\cfrac{R_4}{R_2+R_1||R_{load}}\cfrac{R_1}{R_1+R_{load}}$$
Current from Rload because of V2 from terminal A to terminal B (Fig. 4),
$$I_{V2}=\cfrac{V_2}{R_2+R_4+R_{load}||R_1}\cfrac{R_1}{R_{load}+R_1}$$
Superimpose the calculated currents and voltages together.
The current through Rload is (simply adding all the current obtained above with their sign) :
$$I_{load}=I_{V1}+I_{I1}+I_{V2}$$