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ure 10 V <1 of 3 > Which of the following equations results from correct application of the loop rule for the circuit loop containing R₁, R₂, and Rs? Refer to this diagram (Figure 3) to answer the question ▸ View Available Hint(s) O IRs+IR₂ +10 V+IR₂=0 O IRs-IR₁+10 V-IR₂=0 O IR-IR₁-10 V-I₁R₂=0 O -IsRs-IR +10 V-IR₂=0 O IR-IR₁-10 V+I₁R₂=0 Submit Part C Complete previous part(s) SOLVE Now that you've set up the problem, choose appropriate equations and solve for your unknowns. Part D Complete previous part(s) REFLECT Think about whether your results make sense Part E Complete previous part(s) Feedback

ure 10 V <1 of 3 > Which of the following equations results from correct application of the loop rule for the circuit loop containing R₁, R₂, and Rs? Refer to this diagram (Figure 3) to answer the question ▸ View Available Hint(s) O IRs+IR₂ +10 V+IR₂=0 O IRs-IR₁+10 V-IR₂=0 O IR-IR₁-10 V-I₁R₂=0 O -IsRs-IR +10 V-IR₂=0 O IR-IR₁-10 V+I₁R₂=0 Submit Part C Complete previous part(s) SOLVE Now that you've set up the problem, choose appropriate equations and solve for your unknowns. Part D Complete previous part(s) REFLECT Think about whether your results make sense Part E Complete previous part(s) Feedback

College Physics
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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Figure R₂ 10 V + 5V + R5 R₁ 1 of 3 > Part B Which of the following equations results from correct application of the loop rule for the circuit loop containing R₁, R₂, and Rs? Refer to this diagram (Figure 3) to answer the question. ▸ View Available Hint(s) O IR + I R₁ +10 V+I₁R₂=0 O IR-IR₁ +10 V-IR₂ = 0 O IR-I₁R₁-10 V-I₁ R₂ = 0 O-IsRs-IR₁ +10 V-IR₂0 O IR-IR₁ - 10 V + I R₂=0 Submit Part C Complete previous part(s) SOLVE Now that you've set up the problem, choose appropriate equations and solve for your unknowns. Part D Complete previous part(s) REFLECT Think about whether your results make sense. Part E Complete previous part(s) Provide Feedback
Transcribed Image Text:Figure R₂ 10 V + 5V + R5 R₁ 1 of 3 > Part B Which of the following equations results from correct application of the loop rule for the circuit loop containing R₁, R₂, and Rs? Refer to this diagram (Figure 3) to answer the question. ▸ View Available Hint(s) O IR + I R₁ +10 V+I₁R₂=0 O IR-IR₁ +10 V-IR₂ = 0 O IR-I₁R₁-10 V-I₁ R₂ = 0 O-IsRs-IR₁ +10 V-IR₂0 O IR-IR₁ - 10 V + I R₂=0 Submit Part C Complete previous part(s) SOLVE Now that you've set up the problem, choose appropriate equations and solve for your unknowns. Part D Complete previous part(s) REFLECT Think about whether your results make sense. Part E Complete previous part(s) Provide Feedback
PSS 19.2: Kirchhoff's Rules Learning Goal: To practice Problem-Solving Strategy 19.2 Kirchhoff's rules. In the circuit shown (Figure 1), find the current in each branch. Figure R₂ 10 V 5V + R₂ R₂ R₁ 1 of 3 > Problem-solving Strategy 19.2 Kirchhoff's rules 1. Draw a large circuit diagram so that you have plenty of room for labels. Label all quantities, known and unknown, including an assumed direction for each unknown current and emf. If the actual direction of a particular quantity is opposite to your assumed direction, the result will come out with a negative sign 2. Usually, when you label currents, it is best to use the junction rule immediately to express the currents in terms of as few quantities as possible. BOLVE 3. Choose any closed loop in the network, and designate a direction (clockwise or counterclockwise) to go around the loop when applying the loop rule. The direction doesn't have to be the same as any assumed current's direction. 4. Go around the loop in the designated direction, adding potential differences as you cross them. An emf is counted as positive when you traverse it from - to + and negative when you traverse it from + to - An IR product is negative if your path passes through the resistor in the same direction as the assumed current and positive if it passes through in the opposite direction. "Uphill potential changes are always positive; "downhill" changes are always negative. REFLECT 9. Always remember that when you go around a loop, adding potential differences in accordance with Kirchhoff's loop rule, rises in potential are positive and drops in potential are negative. 5. Apply Kirchhoff's loop rule to the potential differences obtained in Step 4: V = 0. 6. If necessary, choose another loop to get a different relation among the unknowns, and continue until you have as many independent equations as unknowns or until every circuit element has been included in at least one of the chosen loops. 7. Finally, solve the equations, by substitution or some other means, to determine the unknowns. 8. You can use this same bookkeeping system to find the potential VA of any point A with respect to any other point B. Start at B and add the potential changes that you encounter in going from B to A using the same sign rules as in Step 4. The algebraic sum of these changes is VAB VA - VB. SET UP Before writing any equations, organize your information and draw appropriate diagrams. ▾ Part A <3 of 12 The three currents in the circuit are defined in the figure. (Figure 2) Which of the following equations results from correct application of the junction rule for I₁, I₂, and I3? ▸ View Available Hint(s) Review | Constar O I +12 + ₁ = 0 O I₁+I2 Is=0 O I₁-12 + Is=0 O I-12-13 0 Submit ▾ Part B
Transcribed Image Text:PSS 19.2: Kirchhoff's Rules Learning Goal: To practice Problem-Solving Strategy 19.2 Kirchhoff's rules. In the circuit shown (Figure 1), find the current in each branch. Figure R₂ 10 V 5V + R₂ R₂ R₁ 1 of 3 > Problem-solving Strategy 19.2 Kirchhoff's rules 1. Draw a large circuit diagram so that you have plenty of room for labels. Label all quantities, known and unknown, including an assumed direction for each unknown current and emf. If the actual direction of a particular quantity is opposite to your assumed direction, the result will come out with a negative sign 2. Usually, when you label currents, it is best to use the junction rule immediately to express the currents in terms of as few quantities as possible. BOLVE 3. Choose any closed loop in the network, and designate a direction (clockwise or counterclockwise) to go around the loop when applying the loop rule. The direction doesn't have to be the same as any assumed current's direction. 4. Go around the loop in the designated direction, adding potential differences as you cross them. An emf is counted as positive when you traverse it from - to + and negative when you traverse it from + to - An IR product is negative if your path passes through the resistor in the same direction as the assumed current and positive if it passes through in the opposite direction. "Uphill potential changes are always positive; "downhill" changes are always negative. REFLECT 9. Always remember that when you go around a loop, adding potential differences in accordance with Kirchhoff's loop rule, rises in potential are positive and drops in potential are negative. 5. Apply Kirchhoff's loop rule to the potential differences obtained in Step 4: V = 0. 6. If necessary, choose another loop to get a different relation among the unknowns, and continue until you have as many independent equations as unknowns or until every circuit element has been included in at least one of the chosen loops. 7. Finally, solve the equations, by substitution or some other means, to determine the unknowns. 8. You can use this same bookkeeping system to find the potential VA of any point A with respect to any other point B. Start at B and add the potential changes that you encounter in going from B to A using the same sign rules as in Step 4. The algebraic sum of these changes is VAB VA - VB. SET UP Before writing any equations, organize your information and draw appropriate diagrams. ▾ Part A <3 of 12 The three currents in the circuit are defined in the figure. (Figure 2) Which of the following equations results from correct application of the junction rule for I₁, I₂, and I3? ▸ View Available Hint(s) Review | Constar O I +12 + ₁ = 0 O I₁+I2 Is=0 O I₁-12 + Is=0 O I-12-13 0 Submit ▾ Part B
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