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.
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Answer:
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Explanation:
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