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1. Use superposition to calculate the magnitudes of all currents in the circuit shown below (show all work):

Magnitudes of the currents are i1= 0.00104A , i2= 0.003641A , i3= 0.00508A.

Explanation:

Using superposition theorem,

remove the E1 voltage supply source and calculate resistance across it.

From the circuit given, as the resistances R1 and R2 are parallel

using the formula R1//R2=(R1.R2/(R1+R3)

V1= ((r1||r2)/(r1+r2||r3))*E1

v1 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*10v

v1= 2.3v

v2 = ((r1||r2)/(r1+r2||r3))*E2

v2 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*5v

v2= 1.161v

V = V1+V2

=> 2.3 + 1.161

=> 3.461v

Magnitudes of the currents can be found by i=v/r

i1 = v/r1

=> 3.461/3.3kΩ

=>0.00104A

i2=2.89/1kΩ

=>0.003461A

i3=2.89/680Ω

=> 0.00508A.

Therefore the magnitudes of the currents are i1, i2, and i3.

answer:

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