Kirchhoff's Law Problem -- Resistors in Series and Parallel and a Switch

AI Thread Summary
The discussion revolves around solving a problem involving Kirchhoff's Law, specifically with resistors in series and parallel configurations. Participants clarify the correct equations for calculating total resistance, emphasizing the importance of separating resistors into distinct groups. There is confusion regarding the values of resistances and the application of Kirchhoff's Voltage Law (KVL) to validate the calculations. The conversation also touches on the nature of a specific resistor (R7) and its changing resistance based on the switch's position. Overall, the thread highlights the collaborative effort to resolve misunderstandings in circuit analysis.
DevonZA
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Homework Statement


upload_2016-4-22_13-22-17.png


Homework Equations


Rparellel=1/R1+1/R2...
IR1=R2/R1+R2 x I

The Attempt at a Solution


upload_2016-4-22_13-22-43.png


I am unsure of how to answer d) and e) using KVL because I count 4 junctions?
Where should I start?
 
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Check your first equation ! It's not for Rpalrallel but for 1/=Rpalrallel !

And the 20 and 25 Ohm resistors are NOT parallel to e.g. the 12 Ohm !
 
Yes sorry that was a typo.
The total resistance for the parallel resistors still equals 15.18ohms but yes I should have done them separately:
(1/12+1/10+1/16)^-1 = 4.07ohms
(1/20+1/25)^-1= 11.11ohms

Do my values look correct?
 
I only know $${1\over 12} + {1\over 10}+ {1\over 16}+ {1\over 20}+ {1\over 25} \ne 15.82$$
Your values in brackets are probably the book answers ?
 
(1/12+1/10+1/16+1/20+1/25)^-1=15.18 ohms

Yes values are answers in the book
 
Then why ask if the values are correct ?
 
DevonZA said:
(1/12+1/10+1/16+1/20+1/25)^-1=15.18 ohms
No. $${1\over 12} + {1\over 10}+ {1\over 16}+ {1\over 20}+ {1\over 25} = {1\over 2.98}$$However,$$
{1\over {1\over 12} + {1\over 10}+ {1\over 16} } + {1 \over {1\over 20}+ {1\over 25} } = 4.07 + 11.11 = 15.18$$
 
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upload_2016-4-22_14-27-51.png
 
DevonZA said:

Homework Statement


[ ATTACH=full]99511[/ATTACH]
upload_2016-4-22_13-22-17-png.99511.png
What sort of device is R7 that its resistance changes depending on whether the switch is opened or closed
 
  • #10
The parts d and e are a bit weird, because basically you use KVL to calculate R7 in a and b.
And it wouldn't be proof, just showing.
I should think they want you to show that 1.2 A * (4.07 + 60 + 11.11 + 24.82) Ohm = 120 V
 
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  • #11
+1

Although I prefer to make the voltages around the loop sum to zero.
 
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  • #12
@DevonZA ,

Are you saying that this problem is from a textbook?

What book? Who is the author ?
 
Last edited:
  • #13
CWatters said:
+1

Although I prefer to make the voltages around the loop sum to zero.
Can you show this?
 
  • #14
SammyS said:
@DevonZA ,

Are you saying that this problem is from a textbook?

What book? Who is the author ?
Hi Sammy this is from our varsity study guide. The author is most likely the lecturer who doesn't respond to me hence why I must ask these questions on here.
 
  • #15
CWatters said:
Although I prefer to make the voltages around the loop sum to zero.
DevonZA said:
Can you show this?

With reference to the equivalent circuit below... KVL says that going around a loop the voltages sum to zero.
KVL.png


If I arbitrarily choose to start at the -ve terminal of the battery and go around clockwise we have to prove that..

+120 + (-V1) + (-V2) + (-V3) + (-V4) = 0

V1 = 1.2 * 4.07 = 4.884V
V2 = 1.2 * 60 = 72V
V3 = 1.2 * 11.11 = 13.332V
V4 = 1.2 * 24.82 = 29.784V

Substitute..

+120 - 4.884 - 72 -13.332 - 29.784 = 0
 
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  • #16
CWatters said:
With reference to the equivalent circuit below... KVL says that going around a loop the voltages sum to zero.
View attachment 99596

If I arbitrarily choose to start at the -ve terminal of the battery and go around clockwise we have to prove that..

+120 + (-V1) + (-V2) + (-V3) + (-V4) = 0

V1 = 1.2 * 4.07 = 4.884V
V2 = 1.2 * 60 = 72V
V3 = 1.2 * 11.11 = 13.332V
V4 = 1.2 * 24.82 = 29.784V

Substitute..

+120 - 4.884 - 72 -13.332 - 29.784 = 0
Thank you for the nice clear explanation CWatters
 
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