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The circuit Of Figure 7.7 has:
vs = 10cos2t V
Is = 3cos(3t + π/4) A
R1 = 5 Ω
R2 = 10 Ω
Use the superposition theorem to calculate the current i through R2.
The strings:
S7P4A41 (motion - linear).
The math:
Pj Problem of Interest is of type motion (linear).
Superposition Theorem :in a linear network containing multiple sources, the voltage across or current through any passive element may be found as the algebraic sum of the individual voltages or currents due to each of the independent sources acting alone, with all other independent sources deactivated.
An ideal voltage source is deactivated by replacing it with a short circuit.
An ideal current source is deactivated by replacing it with an open circuit.
Deactivate the current source
So current source (Is) loop is open circuit
Let the resulting current through R2 be i' given this scenario
So, i' = Vs/(R1 + R2) = 10cost2t/15 = (2/3)cos2t = 0.667cos2t.
Deactivate the voltage source
So, voltage source (Vs) is replaced with a short circuit
Let the resulting current through R2 be i'' and the voltage response be v'' given this scenario
So, R1 || R2 = Req = 50/15 = 10/3
v'' = (10/3)Is = R2i''
So, i'' = [(10/3)3cos(3t + π/4)/10] = cos(3t + π/4)
So, by the superposition theorem:
i = i' + i'' = 0.667cos2t + cos(3t + π/4) A.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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