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iPad? 2:05 AM 84% lon-capa.auamed.net Sea Demonstra. Amazon.c. Gaston Br.. seated axl. Hot New C.. Crest 3D... LON-CA... Ashley Forde (Student) Physics 2-Spring 2018 Messages Courses Help Logou Main Menu | Course Contents Course Contents... Chapter 21 Kirchhoff 2 ?Notes 11 Bookmark à Evaluate ecommunicate Print Find the currents flowing in the circuit in the figure below, Use the direction of the currents as shown in the figure. Explicitly show how you follkw the steps in the Problem-Solving Strategies for Series and Parallel Resistors. (Hint: For resistors in series, you may want to add the resistances since they have a common current. 12 ? 0.5 ? 22V 4.oV 05? 025 ? 26V ? 14 ? 075? seTries 0/10 eTries 0/10 Tries 0/10 Post Discussion Send Feedback

Explanation / Answer

The problem can be solved using Kirchhoff's Laws.

At the junction b, per Kirchhoff's junction rule,

I1=I2+I3 ...(i)

Let us apply Kirchhoff's loop rule in 2 loops : abefa and abcdefa

loop abefa:

22-0.5I1-18I1-8I3+4-0.25I3-14I1=0

Or, 26-32.5I1-8.25I3=0...(ii)

loop abdcefa

22-0.5I1-18I1-12I2+14-0.5I2-0.75I2 -26-14I1=0

Or, 10-32.5I1-13.25I2=0...(iii)

solving (i),(ii) and (iii), we get

I1=0.528 A

I2=-0.541 A

I3=1.07 A

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