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6.70 A D Solar Panel 0.385 Ω ww Battery M 0.385 Ω 0.77 Ω ww 0.028 02 12.40 V KHM LED Light MA 0.77 Ω 0.77 Ω www 7.20 -5.50 V LED Light M 0.77 Ω 0.77 Ω ww 7.20 5.50 V LED Light at gate www 0.77 Ω ww 7.20 75.50 V Re-draw the full circuit diagram of the system described above. These are the actual values that should be used in your calculations. Where possible, simplify and redraw the diagram by combining series and parallel resistances.

6.70 A D Solar Panel 0.385 Ω ww Battery M 0.385 Ω 0.77 Ω ww 0.028 02 12.40 V KHM LED Light MA 0.77 Ω 0.77 Ω www 7.20 -5.50 V LED Light M 0.77 Ω 0.77 Ω ww 7.20 5.50 V LED Light at gate www 0.77 Ω ww 7.20 75.50 V Re-draw the full circuit diagram of the system described above. These are the actual values that should be used in your calculations. Where possible, simplify and redraw the diagram by combining series and parallel resistances.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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6.70 A/ ↑ Solar Panel 0.385 Ω M M Battery M 0.385 Ω 0.77 Ω ww 0.028 02 12.40 V MD LED Light M 0.77 Ω ww 7.2 Q 0.77 Ω -5.50 V LED Light www 0.77 Ω 0.77 Ω ww 7.2 Q 5.50 V LED Light at gate 0.77 Ω 7.2 Q 75.50 V Re-draw the full circuit diagram of the system described above. These are the actual values that should be used in your calculations. Where possible, simplify and redraw the diagram by combining series and parallel resistances.
Transcribed Image Text:6.70 A/ ↑ Solar Panel 0.385 Ω M M Battery M 0.385 Ω 0.77 Ω ww 0.028 02 12.40 V MD LED Light M 0.77 Ω ww 7.2 Q 0.77 Ω -5.50 V LED Light www 0.77 Ω 0.77 Ω ww 7.2 Q 5.50 V LED Light at gate 0.77 Ω 7.2 Q 75.50 V Re-draw the full circuit diagram of the system described above. These are the actual values that should be used in your calculations. Where possible, simplify and redraw the diagram by combining series and parallel resistances.
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with this simplified circuit, i'm not 100% sure that it's right, because where did my current source go? how can it combine with my voltage source? please help. 

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when finding thevenin's equivalent voltage, why is 6.70 A multiplied by 0.028 Ohms, the resistance of the voltage source of 12.40V? could you explain why and how that is the thevenin formula that it relates to? 

also, I don't understand why the simplified resistance equivalence of 1.568 Ohms was put there. wouldn't it be on the next resistor over in the circuit? 

what do the nodes of A & B mean in that final simplified circuit? 

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