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

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SUMMARY

The discussion focuses on solving a circuit problem involving Kirchhoff's Voltage Law (KVL) and resistors in series and parallel configurations. Participants clarify the calculations for total resistance, specifically addressing the parallel resistors of 12 Ohm, 10 Ohm, 16 Ohm, 20 Ohm, and 25 Ohm, confirming that the total resistance is 15.18 Ohms. The conversation also emphasizes the importance of correctly applying KVL to verify voltage sums around a loop, with specific voltage calculations provided for each resistor. The problem originates from a varsity study guide, indicating a practical application of theoretical concepts.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Knowledge of resistor configurations: series and parallel
  • Ability to perform calculations involving Ohm's Law
  • Familiarity with basic circuit analysis techniques
NEXT STEPS
  • Study advanced applications of Kirchhoff's Laws in complex circuits
  • Learn about Thevenin's and Norton's Theorems for circuit simplification
  • Explore the effects of varying resistor values on circuit behavior
  • Investigate the use of simulation tools like LTspice for circuit analysis
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in practical electronics who seeks to deepen their understanding of circuit analysis and resistor configurations.

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,$$
{1\over {1\over 12} + {1\over 10}+ {1\over 16} } + {1 \over {1\over 20}+ {1\over 25} } = 4.07 + 11.11 = 15.18$$
 
Last edited:
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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