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R₁ 5.02 R₂ 7892 T4 0.1002 T₂ = 48.0 V 0.500 £, -6.0 V 5 mi i 0.050 h R₂ 400 R₂ 200 44 20.20 Ω 2 -36.0V (a) Apply the junction rule at point a. Use the convention that current leaving a junction is positive and current entering a junction is negative. 4₁x| +1₂x[ |+1₂x[ The next two parts of the problem will require you to use the math type box to enter an equation for Kirchhoff's loop rule in terms of variables only. An example problen with what the required formatting should look like for the system to recognize it as a correct answer is provided below. ✓=0 Example: Apply the loop rule to loop abcdelka, using variables instead of values. (i.e. E₁, 12, R₁, ₂ etc.). Use the Mathtype Editor (orange box). Use subscripts using the menu in the math editor. The answer is case sensitive (use R and r differently as appropriate). When forming the product IR, don't forget the multiplication sign. Enter IR as I-R. Solution: -1₁R₁+E₁-₁₁-1₁ R₁+1/₂ R₂+1/₂ √₂-E₂=0 (b) Apply the loop rule to loop abcdefghija, using variables instead of values. (i.e. E₁, 12, R₁, 2 etc.). Use the Mathtype Editor (orange box). =0 (c) Apply the loop rule to loop akledcba, using variables instead of values (ie. E₁, 12, R₁, 2 etc.) Use

R₁ 5.02 R₂ 7892 T4 0.1002 T₂ = 48.0 V 0.500 £, -6.0 V 5 mi i 0.050 h R₂ 400 R₂ 200 44 20.20 Ω 2 -36.0V (a) Apply the junction rule at point a. Use the convention that current leaving a junction is positive and current entering a junction is negative. 4₁x| +1₂x[ |+1₂x[ The next two parts of the problem will require you to use the math type box to enter an equation for Kirchhoff's loop rule in terms of variables only. An example problen with what the required formatting should look like for the system to recognize it as a correct answer is provided below. ✓=0 Example: Apply the loop rule to loop abcdelka, using variables instead of values. (i.e. E₁, 12, R₁, ₂ etc.). Use the Mathtype Editor (orange box). Use subscripts using the menu in the math editor. The answer is case sensitive (use R and r differently as appropriate). When forming the product IR, don't forget the multiplication sign. Enter IR as I-R. Solution: -1₁R₁+E₁-₁₁-1₁ R₁+1/₂ R₂+1/₂ √₂-E₂=0 (b) Apply the loop rule to loop abcdefghija, using variables instead of values. (i.e. E₁, 12, R₁, 2 etc.). Use the Mathtype Editor (orange box). =0 (c) Apply the loop rule to loop akledcba, using variables instead of values (ie. E₁, 12, R₁, 2 etc.) Use

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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R₁ 5.02 R₂ 7892 £₁ = 24.0 V +P+ 0.1002 ₂= 48.0 V HA 0.50 Ω £,- 6.0V 5 i 0.05 02 h R₂ www 40 Q R₁ 200 ₁-36.0 V (a) Apply the junction rule at point a. Use the convention that current leaving a junction is positive and current entering a junction is negative. 1₁x ✓+1₂x | + 13×[ The next two parts of the problem will require you to use the math type box to enter an equation for Kirchhoff's loop rule in terms of variables only. An example problen with what the required formatting should look like for the system to recognize it as a correct answer is provided below. Example: Apply the loop rule to loop abcdelka, using variables instead of values. (i.e. E₁, 12, R₁, r₂ etc.). Use the Mathtype Editor (orange box). (d) Find the currents flowing in the circuit. 4₁ = • Use subscripts using the menu in the math editor. • The answer is case sensitive (use R and r differently as appropriate). • When forming the product IR, don't forget the multiplication sign. Enter /R as I-R. Solution: -1₁R₁+E₁₁₁-1₁° R5 +1₂ R₂ +1₂-1₂-E₂ = 0 (b) Apply the loop rule to loop abcdefghija, using variables instead of values. (i.e. E₁, 12, R₁, 12 etc.). Use the Mathtype Editor (orange box). =0 (c) Apply the loop rule to loop akledcba, using variables instead of values (i.e. E₁, 12, R₁, 12 etc.) Use the mathtype editor (orange box). ✓=0 = 0 13² A Note: The sign convention for voltage drops when using Kirchhoff's loop rule are: • When a resistor is traversed in the same direction as the current, the change in potential is -IR When a resistor is traversed in the direction opposite to the current, the change in potential is +IR • When an emf is traversed from - to + (the same direction it moves positive charge), the change in potential is +emf. • When an emf is traversed from + to-(opposite to the direction it moves positive charge), the change in potential is -emf.
Transcribed Image Text:R₁ 5.02 R₂ 7892 £₁ = 24.0 V +P+ 0.1002 ₂= 48.0 V HA 0.50 Ω £,- 6.0V 5 i 0.05 02 h R₂ www 40 Q R₁ 200 ₁-36.0 V (a) Apply the junction rule at point a. Use the convention that current leaving a junction is positive and current entering a junction is negative. 1₁x ✓+1₂x | + 13×[ The next two parts of the problem will require you to use the math type box to enter an equation for Kirchhoff's loop rule in terms of variables only. An example problen with what the required formatting should look like for the system to recognize it as a correct answer is provided below. Example: Apply the loop rule to loop abcdelka, using variables instead of values. (i.e. E₁, 12, R₁, r₂ etc.). Use the Mathtype Editor (orange box). (d) Find the currents flowing in the circuit. 4₁ = • Use subscripts using the menu in the math editor. • The answer is case sensitive (use R and r differently as appropriate). • When forming the product IR, don't forget the multiplication sign. Enter /R as I-R. Solution: -1₁R₁+E₁₁₁-1₁° R5 +1₂ R₂ +1₂-1₂-E₂ = 0 (b) Apply the loop rule to loop abcdefghija, using variables instead of values. (i.e. E₁, 12, R₁, 12 etc.). Use the Mathtype Editor (orange box). =0 (c) Apply the loop rule to loop akledcba, using variables instead of values (i.e. E₁, 12, R₁, 12 etc.) Use the mathtype editor (orange box). ✓=0 = 0 13² A Note: The sign convention for voltage drops when using Kirchhoff's loop rule are: • When a resistor is traversed in the same direction as the current, the change in potential is -IR When a resistor is traversed in the direction opposite to the current, the change in potential is +IR • When an emf is traversed from - to + (the same direction it moves positive charge), the change in potential is +emf. • When an emf is traversed from + to-(opposite to the direction it moves positive charge), the change in potential is -emf.
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