For the left half of the circuit, starting in the bottom left corner and moving clockwise, have AV-AV 0=AVE,+AV+AVI₂+AVE₂ = +₁ − iş R₁ − izR3 - E₂. For the right half of the circuit, starting in the bottom right corner and moving clockwise, have AVioon-AV 0-AV₂ +AV₂ +AV =+2+13R3-1₂R₂. Applying Kirchhoff's junction rule at the top junction, we also have i=₂+is. We now have three equations and three unknowns, which we can solve for the three curren We get -0.030 A -0.050 A ix=-0.020 A.
For the left half of the circuit, starting in the bottom left corner and moving clockwise, have AV-AV 0=AVE,+AV+AVI₂+AVE₂ = +₁ − iş R₁ − izR3 - E₂. For the right half of the circuit, starting in the bottom right corner and moving clockwise, have AVioon-AV 0-AV₂ +AV₂ +AV =+2+13R3-1₂R₂. Applying Kirchhoff's junction rule at the top junction, we also have i=₂+is. We now have three equations and three unknowns, which we can solve for the three curren We get -0.030 A -0.050 A ix=-0.020 A.
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Please explain the algebra behind finding i1, i2, and i3 from the included circuit emfs and resistors.
![For the left half of the circuit, starting in the bottom left corner and moving clockwise, we
have
AVioop - AV
0=AVE₁ +AVR₂ +AVR₂ +AV₂
= +₁ − i₁R₁ − izR3 - E2-
For the right half of the circuit, starting in the bottom right corner and moving clockwise, we
have
AVioop-AV₂
0=AV₂ +AVR₂ +AVR₂
= +2+ 13R3-1₂R₂.
Applying Kirchhoff's junction rule at the top junction, we also have
i=i₂+ią.
We now have three equations and three unknowns, which we can solve for the three currents.
We get
i = 0.030 A
in = 0.050 A
is = -0.020 A.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3600976f-5a74-4553-80f1-af12fd18d682%2Fea33d150-f7e9-419f-a39a-5b7928915ddc%2Fqqevjyo_processed.png&w=3840&q=75)
Transcribed Image Text:For the left half of the circuit, starting in the bottom left corner and moving clockwise, we
have
AVioop - AV
0=AVE₁ +AVR₂ +AVR₂ +AV₂
= +₁ − i₁R₁ − izR3 - E2-
For the right half of the circuit, starting in the bottom right corner and moving clockwise, we
have
AVioop-AV₂
0=AV₂ +AVR₂ +AVR₂
= +2+ 13R3-1₂R₂.
Applying Kirchhoff's junction rule at the top junction, we also have
i=i₂+ią.
We now have three equations and three unknowns, which we can solve for the three currents.
We get
i = 0.030 A
in = 0.050 A
is = -0.020 A.
Expert Solution
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Step 1
Here we have 3 equations and three variables, solving the equation we can find the variables i.e current.
Note:- as you have not provided the values of resistors and potential of batteries I am attaching final values in terms of values of resistors and potential values. Putting given values you can easily find the final answer.
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