₁x R₁ 5.0 22 ● H 1/ a ● R3 78 92 £₁ = 24.0 V V d 0.10 2 E₂ = 48.0 V 0.50 Ω E3= 6.0 V 13 + 1₂x i 0.05 02 h R₂ www 40 Ω (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. ✓ + 13 X Rs 20 Ω e $0.20 $2 Ω E = 36.0 V 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). ✓ = 0
₁x R₁ 5.0 22 ● H 1/ a ● R3 78 92 £₁ = 24.0 V V d 0.10 2 E₂ = 48.0 V 0.50 Ω E3= 6.0 V 13 + 1₂x i 0.05 02 h R₂ www 40 Ω (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. ✓ + 13 X Rs 20 Ω e $0.20 $2 Ω E = 36.0 V 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). ✓ = 0
Related questions
Question
Transcribed Image Text:R₁
5.0 22
H
4/
a
●
R3
78 92
●
£₁ = 24.0 V
V
0.10 2
£₂ = 48.0 V
0.50 Ω
E3= 6.0 V
13
+ 1₂x
i 0.05 02 h
R₂
ww
40 Ω
(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
Rs
-2002
✓+13 X
e
-1₁R₁+E₁-₁₁-1₁ R5+12 R₂+12₂=E₂ = 0
0.20
E = 36.0 V
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
Transcribed Image Text:(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
(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
(d) Find the currents flowing in the circuit.
4₁=
A
12
13=
A
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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