Superposition principle in electronics1/13/2024 ![]() Exercise 2įind the current and voltage across the 12 Ω resistor using the superposition theorem. Again it is necessary to carefully observe the directions of the currents. ![]() Important: in the case of the 7500 Ω resistor, note that the currents add up, because in both circuits they circulate in the same direction when passing through this resistance. The theorem is then applied to find the current through the 7500 Ω resistor: This same current goes through the 1500 Ω and 600 Ω resistors equally, since they are all in series. Important: for this resistance, the currents are subtracted, since they circulate in the opposite direction, according to a careful observation of the figures, in which the directions of the currents have different colors. Now the superposition theorem is applied for each resistance, starting with the 400 Ω: I 2500 Ω = 2 mA - 0.5 mA = 1.5 mA Application of the superposition theorem While the one that passes through the 2500 Ω resistor is: It is necessary to find the equivalent resistance between both, knowing that:ġ / R eq = (1/7500) + (1/2500) = 1/1875 → R eq = 1875 Ωįor this other circuit, the current that passes through the 7500 Ω resistor is found by substituting values in the current divider equation: Where I x is the current in the resistor R x, R eq symbolizes the equivalent resistance e I T is the total current. The current of 2 mA = 0.002 A is divided between the two resistors in the figure, therefore the equation of the current divider is valid: The resistors in the right mesh are in series and can be replaced by a single one: The voltage source is immediately eliminated, to work only with the current source. This current is the same for all resistors. The equivalent resistance is found by adding the value of each resistance, since they are all in series:Īpplying Ohm's Law V = I.R and clearing the current: ![]() To begin with, the current source is eliminated, with which the circuit is as follows: In the circuit shown in the following figure, find the current through each resistor using the superposition theorem. ![]() The worked examples below clarify the use of the theorem in some simple circuits. Calculate the algebraic sum of all the contributions found in the previous steps. Repeat the two steps described for all other sources. Determine the output, either voltage or current, produced by that single source. Deactivate all independent sources following the instructions given at the beginning, except the one to be analyzed. It is then a linear dependence of voltage and current in a resistance. A linear circuit is one whose response is directly proportional to the input.įor example, Ohm's law applied to an electrical resistance states that V = i.R, where V is the voltage, R is the resistance e i is the current. Linear dependence is decisive for the theorem to apply. This theorem allows us to analyze linear circuits that contain more than one independent source, since it is only necessary to calculate the contribution of each one separately. The superposition theorem, in electrical circuits, establishes that the voltage between two points, or the current through them, is the algebraic sum of the voltages (or currents if it is the case), due to each source, as if each one acted in independently.
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