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Content Tuno bore to coarch 40 Ω 50 2 o MI M Inbox (465)-jsk244@nau.edu n ProblemID=181498761 ▼ • For a battery with current in the positive-to-negative direction: AVbat = -(Figure 2) For a resistor: AVR-IR(Eigure 3) 4. Apply the loop law: Part A Part B Find the current I in the circuit shown in the figure. (Eigure 5) Express your answer in amperes. View Available Hint(s) I= IVE ΑΣΦ Submit Part C X + Provide Feedback w ? A D o ΣΔ΄ - 0 a d h for F a 12 M V * O < a O 4 of 9 Review | Constants. Next > > 11:17 PM X I B

Content Tuno bore to coarch 40 Ω 50 2 o MI M Inbox (465)-jsk244@nau.edu n ProblemID=181498761 ▼ • For a battery with current in the positive-to-negative direction: AVbat = -(Figure 2) For a resistor: AVR-IR(Eigure 3) 4. Apply the loop law: Part A Part B Find the current I in the circuit shown in the figure. (Eigure 5) Express your answer in amperes. View Available Hint(s) I= IVE ΑΣΦ Submit Part C X + Provide Feedback w ? A D o ΣΔ΄ - 0 a d h for F a 12 M V * O < a O 4 of 9 Review | Constants. Next > > 11:17 PM X I B

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

Content
<HW 25-1
Item 4
↓
Mastering Physics: HW 25-1
session.masteringphysics.com/myct/itemView?assignmentProblemID=181498761
Learning Goal:
To practice Tactics Box 25.1 Using Kirchhoff's loop law.
Circuit analysis is based on Kirchhoff's laws, which can be
summarized as follows:
• Kirchhoff's junction law says that the total current into
a junction must equal the total current leaving the
junction.
• Kirchhoff's loop law says that if we add all of the
potential differences around the loop formed by a
circuit, the sum of these potential differences must be
zero
While Kirchhoff's junction law is needed only when there are one or
Figure
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• For a battery with current in the positive-to-negative direction: AV bat
• For a resistor: AVR-IR. (Figure 3)
4. Apply the loop law:
Part A
Part B
Find the current I in the circuit shown in the figure. (Figure 5)
Express your answer in amperes.
► View Available Hint(s)
I =
IVE ΑΣΦ
Submit
Part C
Provide Feedback
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Write down the Kirchhoff Junction equation and solve it for I, in terms of I, and I3. Write the result here: I1 Write down the Kirchhoff Loop equation for a loop that starts at the lower left corner and follows the perimeter of the circuit diagram clockwise. 0 = Write down the Kirchhoff Loop equation for a loop that starts at the lower left corner and touches the components 4V, R2, and R3. 0 = The resistors in the circuit have the following values: R1 = 13 N R2 = 6 N R3 = 2 N Solve for all the following (some answers may be negative): I1 Amperes I, = Amperes I3 = Amperes NOTE: For the equations, put in resistances and currents SYMBOLICALLY using variables like R1,R2,R3 and I,I2,I3. Use numerical values of 10 and 4 for the voltages.
Select the correct value for each of the quantities listed below. A capacitor of capacitance C= 2.0 µF is initially charged with charge Qo = 6.0 µC. The %3D capacitor is then discharged by connecting it to a resistor of resistance R = 5.0 MN. V V. • The initial voltage of this capacitor is [Select ] At time t = 20 s: • the charge on the capacitor is Q = [Select ] µC, • the voltage across the capacitor is AV= [Select] Y V, • the current in the circuit is I = [ Select ] [ Select ] and the energy stored in the capacitor is U =