Model the battery and the connecting wires are ideal

Info icon This preview shows pages 67–70. Sign up to view the full content.

View Full Document Right Arrow Icon
Model: The battery and the connecting wires are ideal. Visualize: The figure shows how to simplify the circuit in Figure P32.64 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit to find the current and potential difference of each resistor. Solve: (a) From the last circuit in the figure and from Kirchhoff’s law, 100 V 10 10 A. I = Ω = Thus, the current through the battery is 10 A. Now as we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference. In Step 1 of the above diagram, we return the 10 Ω resistor to the 4 Ω , 4 Ω , and 2 Ω resistors in series. These resistors must have the same 10 A current as the 10 Ω resistance. That is, the current through the 2 Ω and the 4 Ω resistors is 10 A. The potential differences are Δ V 2 = (10 A)(2 Ω ) = 20 V Δ V 4 (left) = (10 A)(4 Ω ) = 40 V Δ V 4 (left) = (10 A)(4 Ω ) = 40 V In Step 2, we return the left 4 Ω resistor to the 20 Ω and 5 Ω resistors in parallel. The two resistors must have the same potential difference Δ V = 40 V. From Ohm’s law, 5 40 V 8 A 5 I = = Ω 20 40 V 2.0 A 20 I = = Ω The currents through the various resistors are I 2 = I 4 = 10 A, I 5 = 8 A, and I 20 = 2.0 A. (b) The power dissipated by the 20 Ω resistor is ( ) ( ) ( ) 2 2 20 20 2.0 A 20 80 W I Ω = Ω = . (c) Starting with zero potential at the grounded point, we travel along the outside path to point a and add/subtract the potential differences on the way: 0 V + (20 Ω ) I 20 + (2 Ω ) I 2 = (20 Ω )(2.0 A) + (2 Ω )(10 A) = 60 V = V a
Image of page 67

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
32.65. Model: The wires and batteries are ideal. Visualize: Solve: Assign currents I 1 , I 2 , and I 3 as shown in the figure. If I 3 turns out to be negative, we’ll know it really flows right to left. Apply Kirchhoff’s loop rule counterclockwise to the top loop from the top right corner: I 1 (5 ) Ω + 12 V – I 3 (10 ) Ω = 0. Apply the loop rule counterclockwise to the bottom loop starting at the lower left corner: I 2 (5 ) Ω + 9 V + I 3 (10 ) Ω = 0. Note that since we went against the current direction through the (10 ) Ω resistor the potential increased. Apply the junction rule to the right middle: I 1 = I 2 + I 3 . These three equations can be solved for the current I 3 by subtracting the second equation from the first then making the substitution I 2 I 1 = – I 3 which was derived from the third equation. The result is I 3 = 3 25 A = 0.12 A, left to right.
Image of page 68
32.66. Model: The wires and batteries are ideal. Visualize: Solve: The circuit has been redrawn for clarity. Assign the currents I 1 , I 2 , and I 3 as shown in the figure. To find the power dissipated by the 2 Ω resistor, we must find the current through it. If I 3 turns out to be negative, we’ll know it really flows bottom to top. Apply Kirchhoff’s loop rule clockwise to the left loop from the bottom left corner: + 12 V – I 1 (4 ) Ω I 3 (2 ) Ω = 0.
Image of page 69

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 70
This is the end of the preview. Sign up to access the rest of the 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