19_InstSolManual_PDF_Part13 - Apply the loop rule to...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
19.58. Set Up: The circuit diagram is given in Figure 19.58. The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents. Figure 19.58 Solve: The loop rule applied to loop (1) gives: The loop rule applied to loop (2) gives: 19.59. Set Up: The circuit is sketched in Figure 19.59. When we go around loop (1) in the direction shown there is a potential rise across the 200.0 V battery, so there must be a drop across R and the current in R must be in the direction shown in the figure. Similar analysis of loops (2) and (3) tell us that currents and must be in the directions shown. The junction rule has been used to label the currents in all the other branches of the circuit. Figure 19.59 Solve: (a)
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Apply the loop rule to loop (1): (b) Loop (2): I 2 5 160.0 V 40.0 V 5 4.00 A 1 160.0 V 2 I 2 1 40.0 V 2 5 0. R 5 1 200.0 V I 1 5 1 200.0 V 10.0 A 5 20.0 V 1 200.0 V 2 I 1 R 5 0. 40.0 V 20.0 V 200.0 V 160 V 1 1 2 1 4 2 1 2 2 1 3 2 R I 1 I 5 I 2 I 1 1 I 2 I 2 1 I 5 + + I 5 I 2 E 2 5 20.0 V 2 1.00 V 2 2.00 V 2 4.00 V 2 6.00 V 5 7.0 V. 1 20.0 V 2 1 1.00 A 21 1.00 V 2 2 1 2.00 A 21 1.00 V 2 2 E 2 2 1 2.00 A 21 2.00 V 2 2 1 1.00 A 21 6.00 V 2 5 E 1 5 20.0 V 2 1.00 V 1 4.00 V 1 1.00 V 2 6.00 V 5 18.0 V. 1 20.0 V 2 1 1.00 A 21 1.00 V 2 1 1 1.00 A 21 4.00 V 2 1 1 1.00 A 21 1.00 V 2 2 E 1 2 1 1.00 A 21 6.00 V 2 5 1.00 V 1.00 V 1.00 V 2.00 V 4.00 V 1.00 A 1.00 A 6.00 V 20.0 V E 1 E 2 2.00 A 1 1 2 + + + Current, Resistance, and Direct-Current Circuits 19-13...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern