![]() To have a physical quantity that is independent of test charge, we define electric potential (or simply potential, since electric is understood) to be the potential energy per unit charge: But we do know that because , the work, and hence , is proportional to the test charge. Calculating the work directly may be difficult, since and the direction and magnitude of can be complex for multiple charges, for odd-shaped objects, and along arbitrary paths. Therefore, although potential energy is perfectly adequate in a gravitational system, it is convenient to define a quantity that allows us to calculate the work on a charge independent of the magnitude of the charge. (The default assumption in the absence of other information is that the test charge is positive.) We briefly defined a field for gravity, but gravity is always attractive, whereas the electric force can be either attractive or repulsive. Recall that earlier we defined electric field to be a quantity independent of the test charge in a given system, which would nonetheless allow us to calculate the force that would result on an arbitrary test charge. Apply conservation of energy to electric systems.Describe systems in which the electron-volt is a useful unit.Calculate electric potential and potential difference from potential energy and electric field.Define electric potential, voltage, and potential difference.On R3 Voltage Drop = (6+3)V = 9V, Current = (0.5+0.25) Amps = 0.By the end of this section, you will be able to: Now finally to find the responses on each branch due to the combined effect of both current source and voltage source we add the individual responses. On R3 Voltage Drop = 3V, Current = 0.25 Amps On R2 Voltage Drop = 3V, Current = 0.5 Amps On R1 Voltage Drop = 3V, Current = 0.25 Amps Thus The responses due to the current source are: To remove the Voltage source it is shorted which converts the circuit into a simple network of parallel and series connection of resistors and the responses can be easily calculated using ohm’s law as following: Superposition Theorem Now let us find the responses on various branches due to the current source: On R3 Voltage Drop = 6V, Current = 0.5 Amps On R2 Voltage Drop = 0V, Current = 0 Amps On R1 Voltage Drop = 6V, Current = 0.5 Amps Thus The responses due to The voltage source are: ![]() To Remove the Current source it is opened, which converts the circuit into a simple voltage divider circuit and the responses can be calculated simply by using ohm’s law as following: Superposition Theorem Let us first Find Responses on the branches due to the Voltage source: We can use Superposition Theorem to solves the circuit as following: In the following circuit: Superposition Theorem Now algebraically add the responses due to each source on a branch to find the response on the branch due to the combined effect of all the sources.Repeat step 2 and 3 for every source the circuit have.Determine the branch responses or voltage drop and current on every branches simply by using KCL, KVL or Ohm’s Law.And replace a Current source with an Open so as to maintain a 0 Amps Current between two terminals of the current source. If the Source is an Ideal source or internal impedance is not given then replace a Voltage source with a short so as to maintain a 0 V potential difference between two terminals of the voltage source. Replace all the voltage and current sources on the circuit except for one of them.While replacing a Voltage source or Current Source replace it with their internal resistance or impedance.First of all make sure the circuit is a linear circuit or a circuit where Ohm’s law implies, because Superposition theorem is applicable only to linear circuits and responses.To solve a circuit with the help of Superposition theorem follow the following steps: ![]() The process of using Superposition Theorem on a circuit: Then the Combined responses (Voltage drop and Current through it) on a branch due to all the sources combined is the algebraic sum of responses on the branches due to each individual sources. If we find the branch responses (Voltage drop and Current through it) on a branch due to only of those source by ignoring effect of all other sources or replacing all other sources by their corresponding internal impedance, and repeat the process for every source on the circuit. ![]() Then Superposition theorem suggests that: Suppose an electrical circuit having several branches and or loads and also several source some being current source and some being voltage source. In a linear circuit with several sources the voltage and current responses in any branch is the algebraic sum of the voltage and current responses due to each source acting independently with all other sources replaced by their internal impedance. ![]()
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