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R₁- 5.02 £₁ = 24.0 V R₂ 78 52 1/1 0.10 E₂= 48.0 V 14 H 1₂= 0.50 Ω 2₂-6.0 V 55 www i 0.050 h d R₂ ww 40 0 Rs 200 (d) Find the currents flowing in the circuit. A e 4 20.20 2 (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 + 1₂ x + 13 x E₁-36.0 V g 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). • 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: = 0 -1₁R₁+E₁-₁-1₁R5+1₂ R₂+1₂ 12-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 (i.e. E₁. 12. R₁. r2 etc.) Use the mathtype editor (orange box). =0

R₁- 5.02 £₁ = 24.0 V R₂ 78 52 1/1 0.10 E₂= 48.0 V 14 H 1₂= 0.50 Ω 2₂-6.0 V 55 www i 0.050 h d R₂ ww 40 0 Rs 200 (d) Find the currents flowing in the circuit. A e 4 20.20 2 (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 + 1₂ x + 13 x E₁-36.0 V g 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). • 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: = 0 -1₁R₁+E₁-₁-1₁R5+1₂ R₂+1₂ 12-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 (i.e. E₁. 12. R₁. r2 etc.) Use the mathtype editor (orange box). =0

College Physics
College Physics
10th Edition
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter18: Direct-Current Circuits
Section: Chapter Questions
Problem 18P: (a) Find the current in each resistor of Figure P18.18 by using the rules for resistors in series...
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R₁ 5.0 Ω a R₂ 78 Ω» T4₁ 1₂ £₁ = 24.0 V 0.102 E2= 48.0 V w 0.50 Ω 2₂= 6.0 V Fa ww i 0.05 02 h R₂ ww 40 Q Rs - 20 Ω e > TA 20.20 Ω E₁36.0 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 + 1₂ x (d) Find the currents flowing in the circuit. 4₁ = А 1₂= SANA А A + 13x 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). • 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: V -₁R₁+E₁-₁₁-1₁ 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, r2 etc.). Use the Mathtype Editor (orange box). √=0 0 (c) Apply the loop rule to loop akledcba, using variables instead of values (i.e. E₁, 12, R₁, 2 etc.) Use the mathtype editor (orange box). √=0 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 5.0 Ω a R₂ 78 Ω» T4₁ 1₂ £₁ = 24.0 V 0.102 E2= 48.0 V w 0.50 Ω 2₂= 6.0 V Fa ww i 0.05 02 h R₂ ww 40 Q Rs - 20 Ω e > TA 20.20 Ω E₁36.0 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 + 1₂ x (d) Find the currents flowing in the circuit. 4₁ = А 1₂= SANA А A + 13x 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). • 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: V -₁R₁+E₁-₁₁-1₁ 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, r2 etc.). Use the Mathtype Editor (orange box). √=0 0 (c) Apply the loop rule to loop akledcba, using variables instead of values (i.e. E₁, 12, R₁, 2 etc.) Use the mathtype editor (orange box). √=0 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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