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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.
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
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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