# KVL in Series-Parallel Circuits

Discussion in 'Homework Help' started by yan500, Jul 17, 2011.

1. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Hey all,

I'm having trouble on Question 6 of this worksheet: http://www.allaboutcircuits.com/worksheets/kvl.html

I keep getting 5.4V as my answer but it is not right; I am around .7V off from the right answer. I've tried just about everything that I know to solve this problem and I can't do it. My total resistance comes out to 3427.37Ω after I combine all the resistors together and I use that number to calculate the voltage drops across the resistors in the path between points A and B.

Am I doing something wrong?

Help is much appreciated!

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,877
1,368
Show as your steps

3. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Alright then,

1. Combine 690 + 1k5 + 1k series resistors into one resistor. This resistor has a value of 3190 ohms.

2. Combine the 4k7 and 250 parallel resistors into one resistor. This resistor has a value of 237.37 ohms.

3. I then combined the two series resistors which made up a total resistance of 3427.37 ohms.

4. From there I made a loop going from point A to B. I used the loop with the 250 and 690 resistors.

5. My KVL equation turned out to be: Vab - (20 *(250/3427.37)) - (20*(690/3427.37)) = 0.

6. Vab = 5.485V

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,877
1,368
You have a error in step 2 I hope that this picture will help you find the error

File size:
17.4 KB
Views:
68
5. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Hmm...from these schematics that you drew I can see that you can ignore the 4k7 resistor. Unfortunately that doesn't solve my problem because if you just add all the series resistors (ignoring the 4k7 resistor) you still get a resistance of 3440 which does not give the right answer.

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,877
1,368
No, you can not ignore the 4k7 resistor.
Step 1
690 + 1k5 + 1k series resistors into one resistor = Rz1 = 3.19K
Step 2
The 4k7 resistor is connect parallel to the Rz1
Rz2 = 4.7K||3.19K
Step3
Rz2 is connect in series withe 250 resistor

Rtotal = Rz2 + 250 = 1.9K + 0.25K = 2.15K

Va = - 20 * 0.25K/2.15K = - 2.325V

and so on

File size:
25.3 KB
Views:
21
7. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Alright so then you have -2.325V + (20* 0.69K/2.15K) = 4.094V. But that's not the correct answer either. The correct answer is 6.148V.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,877
1,368
What is the current that will flow through 690 + 1k5 + 1k=Rz1 ??
If you know the current use Ohm's law to find Vb

And I give you the solution
The voltage across Rz2 =4.7K||3.19K is not equal 20V
VRz2 = 20 - 2.33V = 17.67V
and the Vb voltage
Vb = 17.67V * (0.69K)/3.19K = 3.822V
Vab = -2.33V - 3.822V = -6.15V (-2.3255814 - 3.82299336 = -6.14857476)

9. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
I still don't understand how or where you're pulling these numbers from but I really appreciate you taking the time to help me, thank you.

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,877
1,368
Lets start from the beginning
We have the original circuit (A)

The first step
Rz1 = 690 + 1k5 + 1k = 3.19K

The second step will be replace 4.7K and 3.19K with single resistor.

Rz2 = 4.7K||3.19K = 1.9K

So we end up with figure C. Now lets try write some KVL for circuit in figure C

Vb - V1 - V2 = 0 or V1 = Vb - V2

V2 = It * 250Ω ----> It = 20V/(1.9K+0.25K) = 9.30232558mA

V2 = 2.3255814V

Va voltage is equal -V2 becaues current leaves (flow out) point A.
V1 = Vb - V2 = 20V - 2.3255814V = 17.6744186V
Now we know the V1 voltage so as you can see in figure B
The V1 mus be equal V3 becaues if elements are connect in parallel they will have the same voltage across them.

V1 = V3 = 17.6744186V

If we know V3 without any problem we can find I2

I2 = V3 / Rz1 = 17.6744186V/3.19K = 5.54057009mA

knowing I2 current we easily can find Vb
Vb = I2 * 690Ω = 3.82299336V
The VB voltage is positive becaues current entering point B .
And we assume that voltage is positive where current is enter the resistor.

• ###### 122.PNG
File size:
19.4 KB
Views:
63
Last edited: Mar 6, 2012