MATH1131 - Series & Parallel Circuits

Thread Starter

dennis.muller

Joined Feb 5, 2008
10
Please i need help with my series & parallel circuits math home work, i'm not getting the correct answeres of some values.

Here are the questions & answeres:

MATH1131-01-HANDOUT-020708-Q1-4-PG1-2
MATH1131-01-HANDOUT0-20708-Q1-4-PG2-2

Here are my calculations:

MATH1131-01-HANDOUT-020708-A1-4-PG1-5
MATH1131-01-HANDOUT-020708-A1-4-PG2-5
MATH1131-01-HANDOUT-020708-A1-4-PG3-5
MATH1131-01-HANDOUT-020708-A1-4-PG4-5
MATH1131-01-HANDOUT-020708-A1-4-PG5-5

Here are the above documents in zip format:

MATH1131-HANDOUT-020708-Q1-4-PG9

Please help me identify my errors, Thank you.
 

Attachments

JoeJester

Joined Apr 26, 2005
4,390
From what I've seen of your answers and the answers in the book, your not really in disagreement.

Take answers 1 for example ... The book rounded up the 12.75V to 12.8V, but rounded down the 12.25V to 12.2V They did that so the you wouldn't have that 0.1 V error in KVL when you added them. That doesn't make your answers incorrect ... it just means you went an additional decimal place.

The same can be said for all the answers in the book. How many places should you carry it out? That's up to the instructor. Some would want you to carry it to two decimal places, some want three.

KVL's and KCL's are laws, not suggestions. You can use your knowledge of those two to cross check your answers.

Revisit question 3 ... and apply KVL and KCL.

Revisit question 4 ... and redraw if necessary to allow it to make sense.

Post your results for question 3 before looking at question 4.

There is nothing wrong with your question 1 and 2 results.
 

Thread Starter

dennis.muller

Joined Feb 5, 2008
10
Joe,

Thank you for your help in resolving my math problems. As a visual learner myself, i found it very helpfull re-drawing the circuits as you suggested. I'm still a little fuzzy on what calculations to use when calculating current. Calculating resistance is not a problem for me re-drawing the circuit in such away that the parallel resistors R2 and R3 are considered one resistor which i called R23. But back to calculating the current, i seem to be having problems identifing the correct current to use when solving for the voltage drop accross a resistor or current through a resistor. So i reverted back to re-drawing the circuit and broke it up into three loops, L1(R1,R2), L2(R1,R3), and L3(R4). If I'm correct these are the three posible loops in which current can flow. Purfect so now i have my resistances and current for each loop, this is where i come accross the problem of using the correct current with resistance to calculate voltage drop. After re-doing the question i have concluded that the correct current to use is the loop in which the resistance belongs too. An example would be Resistor 3 (R3), it belongs to Loop 2 (L2). Therefor i would need to calculate the voltage drop on R1 to determin the voltage remaining in Loop 2 (L2) which is V1 = I1 * R1 which equals 3.24V therefor I3 = ((VT-V1)/R3) which equals 225.33µA. Can you suggest alternative methods of calculating or alternative methods of understanding how to calculate?

New Calculations:

MATH1131-01-HANDOUT-020708-A3-PG1-2
MATH1131-01-HANDOUT-020708-A3-PG2-2
MATH1131-01-HANDOUT-020708-A4-PG1-1

Zipped Calculations:

MATH1131-01-HANDOUT-020708-A3-PG9
 

Attachments

Thread Starter

dennis.muller

Joined Feb 5, 2008
10
Joe,

I'v completed re-calculating question 4 as shown below,

MATH1131-01-HANDOUT-020708-A4B-PG1-1

I applyied what i concluded from question 3 in my last post,

After re-doing the question i have concluded that the correct current to use is the loop in which the resistance belongs too.
As a result I labled my calculations by steps 1-7 to illistrate how and why i got my answeres. Steps 1-5 & 7 I found very straigh forward, step 6 is where i had some problems in trying to understand what calculations to use when calculating the voltage drop. What I understand is that the current is divided as it incounters a new loop, how its divided is based on the resistance found in the loop. So as current flows out R1 it divides into R2 & R3 therefor V23=VT-(V1+V2), I2=V23/R2, & I3=V23/R3 and finally V2=I2*R2 & V3=I3*R3. Right now it seems so simple i don't know why i made it so difficult on myself, maybe as a result of making these postings i have given myself a better understanding of the homework questions, KVL, & KCL. Looking back at the questions, the loops I isolated didn't seem of much use to me other then identifying current paths but re-drawing the circuit in series helped me see the circuit differently allowing me to remove the parallel components and replace them in series.
 

JoeJester

Joined Apr 26, 2005
4,390
Dennis,

Back to question 3 you go.


Loop one consists of R1 + ( R2 || R3 ) --- read as R1 in series with the parallel combination of R2 and R3.

Loop two consists of R4

Start at the negative terminal of the battery and draw the current line towards the positive terminal.

It will split at the ( R1 + ( R2 || R3 )) || R4

Then loop one will split at the R2 || R3 and then rejoin at the top node at R1

Then loop 1 and loop 2 rejoin at the top node on it's path to the positive terminal of the battery.

What is suprising is the second page 3 drawing is the first reduction of the parallel circuit.

The voltage drop across that 18.75 ohm resistor is the voltage drop across the resistors that make up that "equvalent resistance". You need to revisit the calculations you made on that first page of question 3.
 

Thread Starter

dennis.muller

Joined Feb 5, 2008
10
Joe,

Thanks for all your help, now that my electrical class is teaching Parallel Circuits everything is clicking into place. I was having so much trouble because these question come from my math class which dosn't give me much insite on circuits. I could use some help on some electrical home work if you search for my postings created by dennis.muller you'll find a new posting "Sine Waves", which i will post soon. Thanks again for your help on the Series & Parallel Circuits.

Dennis Muller
Niagara College
Computer Engineering Technology 1st Year
 
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