Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
19
DC Circuits
Change chapter

19-1: Emf and Terminal Voltage
19-2: Resistors in Series and Parallel
19-3: Kirchhoff's Rules
19-4: Emfs Combined, Battery Charging
19-5: Capacitors in Series and Parallel
19-6: RC Circuits
19-8: Ammeters and Voltmeters

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 34
Q
  1. Determine the currents I1I_1, I2I_2, and I3I_3 in Fig. 19–61. Assume the internal resistance of each battery is r=1.0  Ωr = 1.0 \; \Omega.
  2. What is the terminal voltage of the 6.0-V battery?
Problems 34 and 35.
Figure 19-61.
A
  1. I1=0.43 A, I2=0.31 A, I3=0.11 AI_1 = 0.43 \textrm{ A, }I_2 = 0.31 \textrm{ A, } I_3 = 0.11 \textrm{ A}
  2. 5.9 V5.9 \textrm{ V}
Giancoli 7th Edition, Chapter 19, Problem 34 solution video poster
Padlock

In order to watch this solution you need to have a subscription.

COMMENTS
By ctlawson on Thu, 2/11/2016 - 11:33 AM

Hello,

I think you went over this in previous videos, but does it matter which "loops" you choose? For instance, I chose the top loop and the bottom loop instead of the top loop and larger outer loop, but my values were slightly off. Should I have gotten the same answer? Thanks in advance for your help.

By Mr. Dychko on Fri, 2/12/2016 - 11:34 PM

Hi ctlawson, it doesn't matter which loops you choose. Your goal with 'choosing loops' is to create a collection of equations on which you can then do algebra to figure out the unknowns. You need as many equations as you have unknowns. Each loop is an example of Conservation of Energy, in that the total gain and loss in potential around a closed loop must equal zero. You begin the loop at some point, travel along, and then return to where you started, but if in doing so you arrived with a different amount of energy than when you started, then some energy must have been created or destroyed along the way, which is a violation of Conservation of Energy. Anyhow, I say all this to try and be convincing that the choice of loop doesn't matter. The discrepancy between our answers could be due to intermediate rounding error, or check that you included the 1.0Ω1.0 \Omega resistors, or something like that.

All the best,
Mr. Dychko

By bmuniz8219 on Fri, 4/15/2016 - 9:16 PM

Hi i have a quick question. If I1 points towards the left how does it in include 12OHMs

By Mr. Dychko on Sat, 4/23/2016 - 3:20 AM

Hi bmuniz8219, that's a good question. Each current "starts" at a junction. While the diagram draws I1I_1 to the left along the very top, it's understood (even though not explicitly drawn) that the current originates at the junction below the 12Ω12\Omega resistor where the I2I_2 and I3I_3 currents also meet. So I1I_1 initially goes up through the 12Ω12\Omega resistor, and THEN, left along the top.

All the best with your studies,
Mr. Dychko

By james.alec21 on Fri, 10/14/2022 - 8:42 PM

Hi Mr Dychko, should the I3 value collected in the 2nd loop's equation add up to -28 and not -27?

By james.alec21 on Fri, 10/14/2022 - 8:59 PM

Additionally when solving for I1 should the addition of the two I1 coefficients not yield 2.5032?

By james.alec21 on Fri, 10/14/2022 - 9:21 PM

Ah nevermind I realized that I was missing the one I1 on the left side. Thank you

Find us on:

Facebook iconTrustpilot icon
Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.