
102809, 12:42 PM

#1

Senior Member
Thread Starter
Join Date: Mar 2007
Location: Rohnert Park, CA
Bikes: Pake track, Soma DoubleCross, LeMond Etape, Maruishi RoadAce 303
Posts: 1,248
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Help with electric circuit problem!
I'm stuck on one hw problem that is part of the extra credit portion. I know that I should be able to do it, but I am just missing something.
Here's the problem:
Quote:
Two resistances, R1 and R2, are connected in series across a 12V battery. The current increases by 0.18 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.11 A when R1 is removed, leaving R2 connected across the battery. Find (a)R1 and (b)R2.

So I figure that there are three equations that I can write:
(i) V/(R1+R2) = I
(ii) V/R1 = I + 0.18A
(iii) V/R2 = I + 0.11A
I've tried to algebraically solve for the individual resistors and I just can't seem to do it. I think I am approaching the problem correctly but perhaps I am doing something wrong partway through.
Thanks for your help.



102809, 12:44 PM

#2

On my TARDIScycle!
Join Date: Jun 2005
Location: Eastside Seattlite Termite Mound
Bikes: Trek 520, Trek Navigator 300, Peugeot Versailles PE10DE
Posts: 3,924
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

resistors in parallel or series?
__________________
Quote:
Originally Posted by coffeecake
 it's pretty well established that Hitler was an *******.




102809, 12:46 PM

#3

Senior Member
Thread Starter
Join Date: Mar 2007
Location: Rohnert Park, CA
Bikes: Pake track, Soma DoubleCross, LeMond Etape, Maruishi RoadAce 303
Posts: 1,248
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Originally Posted by KingTermite
resistors in parallel or series?

Series, as stated in the problem.



102809, 12:50 PM

#4

On my TARDIScycle!
Join Date: Jun 2005
Location: Eastside Seattlite Termite Mound
Bikes: Trek 520, Trek Navigator 300, Peugeot Versailles PE10DE
Posts: 3,924
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Originally Posted by darksiderising
Series, as stated in the problem.

Sorry...duh, I missed that. I wanted to make sure you were setting up the problems correctly.
__________________
Quote:
Originally Posted by coffeecake
 it's pretty well established that Hitler was an *******.




102809, 12:55 PM

#5

On my TARDIScycle!
Join Date: Jun 2005
Location: Eastside Seattlite Termite Mound
Bikes: Trek 520, Trek Navigator 300, Peugeot Versailles PE10DE
Posts: 3,924
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

with those three equations (which look correct to me).
(i) V/(R1+R2) = I
(ii) V/R1 = I + 0.18A
(iii) V/R2 = I + 0.11A
Remembering to sub 12 (12V battery) in for V, you have 3 equations and 3 unknowns.
__________________
Quote:
Originally Posted by coffeecake
 it's pretty well established that Hitler was an *******.




102809, 01:01 PM

#6

Fax Transport Specialist
Join Date: May 2008
Location: chicago burbs
Bikes: '17 giant propel, '07 fuji cross pro, '10 gary fisher xcaliber
Posts: 930
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 24 Post(s)

nevermind...
Last edited by black_box; 102809 at 01:13 PM.



102809, 01:02 PM

#7

On my TARDIScycle!
Join Date: Jun 2005
Location: Eastside Seattlite Termite Mound
Bikes: Trek 520, Trek Navigator 300, Peugeot Versailles PE10DE
Posts: 3,924
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Originally Posted by black_box
When you're ready, highlight:

Don't do his homework for him...just help him get in the right direction.
__________________
Quote:
Originally Posted by coffeecake
 it's pretty well established that Hitler was an *******.




102809, 01:14 PM

#8

Fax Transport Specialist
Join Date: May 2008
Location: chicago burbs
Bikes: '17 giant propel, '07 fuji cross pro, '10 gary fisher xcaliber
Posts: 930
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 24 Post(s)

i figured he'd have to show his work anyway?



102809, 01:30 PM

#9

Senior Member
Join Date: Jul 2008
Location: Virginia
Bikes: CRF150
Posts: 205
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 38 Post(s)

would have been more useful if they'd gave you the total current draw with R1 & R2 in circuit...
__________________
Love one another



102809, 01:55 PM

#10

Senior Member
Join Date: Jul 2005
Location: Saratoga, CA
Bikes:
Posts: 11,606
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 3 Post(s)

Quote:
Originally Posted by ritepath
would have been more useful if they'd gave you the total current draw with R1 & R2 in circuit...

No need for that, he's got sufficient info already to solve the problem.



102809, 02:31 PM

#11

Senior Member
Thread Starter
Join Date: Mar 2007
Location: Rohnert Park, CA
Bikes: Pake track, Soma DoubleCross, LeMond Etape, Maruishi RoadAce 303
Posts: 1,248
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

So I have the equations right, which is what I figured. Now the difficulty is solving for one of the unknown variables (the total resistance, R1, and R2)
(i) V/(R1+R2) = I
(ii) V/R1 = I + 0.18A
(iii) V/R2 = I + 0.11A
V = 12 V
Solving for I (the common variable in each equation), I get:
(i) I = (12 V)/(R1+R2)
(ii) I = [(12 V)/R1]  0.18A
(iii) I = [(12 V)/R2]  0.11A
(i) can be rewritten as I = [(12 V)/R1]  [(12 V)/(66.67 ohms)]
(ii) can be rewritten as I = [(12 V)/R2]  [(12 V)/(109.1 ohms)]
When I set (i) and (ii) equal to each other, I get:
Quote:
R1 = [(171.43 ohms)(R2)] / [(171.43 ohms) + R2]
R2 = [(171.43 ohms)(R1)] / [(171.43 ohms)  R1]

But now what the heck to I do? If I sub in one of them into the other, I just prove that the equations are true and don't end up solving anything.
Last edited by darksiderising; 102809 at 02:39 PM.



102809, 03:20 PM

#12

Senior Member
Join Date: Aug 2008
Location: Laurel, MD
Bikes: '07 Felt F85, '14 Cervelo S3
Posts: 71
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

R1 = 37.4169 ohms
R2 = 47.8639 ohms
When only R1 is connected, I = 0.3207A
When only R2 is connected, I = 0.2507A
When both are in series, I = 0.1407A



102809, 05:53 PM

#13

Senior Member
Thread Starter
Join Date: Mar 2007
Location: Rohnert Park, CA
Bikes: Pake track, Soma DoubleCross, LeMond Etape, Maruishi RoadAce 303
Posts: 1,248
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Originally Posted by rajarajan
R1 = 37.4169 ohms
R2 = 47.8639 ohms
When only R1 is connected, I = 0.3207A
When only R2 is connected, I = 0.2507A
When both are in series, I = 0.1407A

Thanks raj. Could you show me how you did that? I'd like to understand it, too.



102809, 06:24 PM

#14

Senior Member
Join Date: Sep 2006
Bikes: Kona Cinder Cone, Sun EZ3 AX
Posts: 1,195
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Never mind...I'll leave it for rajarajan to explain.
Last edited by deraltekluge; 102809 at 06:29 PM.



102809, 06:55 PM

#15

Dirt Bomb
Join Date: Aug 2006
Location: Illinois
Bikes:
Posts: 2,648
Mentioned: 5 Post(s)
Tagged: 0 Thread(s)
Quoted: 206 Post(s)

Quote:
Originally Posted by rajarajan
R1 = 37.4169 ohms
R2 = 47.8639 ohms
When only R1 is connected, I = 0.3207A
When only R2 is connected, I = 0.2507A
When both are in series, I = 0.1407A

Yea, I'd like to see how you did that.
One other question; When both resistors are in series, you're not getting 12V across each resistor, are you? I thought the total V in the circuit =Vr1 + Vr2.
Last edited by sknhgy; 102909 at 02:25 PM.



102809, 06:59 PM

#16

Senior Member
Thread Starter
Join Date: Mar 2007
Location: Rohnert Park, CA
Bikes: Pake track, Soma DoubleCross, LeMond Etape, Maruishi RoadAce 303
Posts: 1,248
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Originally Posted by sknhgy
One other question; When both resistors are in series, your not getting 12V across each resistor, are you? I thought the total V in the circuit =Vr1 + Vr2.

You're right about the voltage being different across different resistors. The current is the same through both, but the voltage across the entire series is the sum of the voltage across each individual resistor.
V=IR, so the voltage across each resistor is the product of the current and the resistance of the resistor.
In a parallel circuit, the voltage across each resistor in parallel is the same as the voltage across the entire circuit. You can calculate the current through each resistor using I=V/R.
Last edited by darksiderising; 102809 at 07:04 PM.



102809, 07:21 PM

#17

Dirt Bomb
Join Date: Aug 2006
Location: Illinois
Bikes:
Posts: 2,648
Mentioned: 5 Post(s)
Tagged: 0 Thread(s)
Quoted: 206 Post(s)

Quote:
Originally Posted by darksiderising
You're right about the voltage being different across different resistors. The current is the same through both, but the voltage across the entire series is the sum of the voltage across each individual resistor.
V=IR, so the voltage across each resistor is the product of the current and the resistance of the resistor.
In a parallel circuit, the voltage across each resistor in parallel is the same as the voltage across the entire circuit. You can calculate the current through each resistor using I=V/R.

Thanks, but that still don't 'splain how Raj solved for R1 and R2.



102809, 07:55 PM

#18

Senior Member
Thread Starter
Join Date: Mar 2007
Location: Rohnert Park, CA
Bikes: Pake track, Soma DoubleCross, LeMond Etape, Maruishi RoadAce 303
Posts: 1,248
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Originally Posted by sknhgy
Thanks, but that still don't 'splain how Raj solved for R1 and R2.

Haha yeah. If I could explain that I'd be a little happier.



102809, 08:57 PM

#19

Fax Transport Specialist
Join Date: May 2008
Location: chicago burbs
Bikes: '17 giant propel, '07 fuji cross pro, '10 gary fisher xcaliber
Posts: 930
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 24 Post(s)

(i) V/(R1+R2) = I
(ii) V/R1 = I + 0.18A
(iii) V/R2 = I + 0.11A
Replace V with 12 and solve equation (ii) for R1 (the only variables will be R1 and I)
Replace V with 12 and solve equation (iii) for R2 (the only variables will be R2 and I)
Replace V with 12, replace R1 and R2 with the equations from the above steps, now you only have I.
Solve for I, then use equation (ii) to find R1 and equation (iii) to find R2.



102909, 08:46 AM

#20

Senior Member
Join Date: Aug 2008
Location: Laurel, MD
Bikes: '07 Felt F85, '14 Cervelo S3
Posts: 71
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Originally Posted by darksiderising
The current increases by 0.18 A when R2 is removed, leaving R1 connected across the battery.

So 12/R1  12/(R1+R2) = 0.18
which leads to 12R2 = 0.18 * R1 * (R1+R2)  equation (a)
Quote:
Originally Posted by darksiderising
The current increases by just 0.11 A when R1 is removed, leaving R2 connected across the battery.

So 12/R2  12/(R1+R2) = 0.11
which leads to 12R1 = 0.11 * R2 * (R1+R2)  equation (b)
As one can see (R1+R2) common in the two equations, divide (b) by (a) and solve to get
R2*R2 = (R1*R1 * 18 ) / 11
leading to R2 = R1 * sqrt (18/11)  equation (c)
Substitute (c) in (a) to get
R1 * (1+ sqrt (18/11)) = 1200 / sqrt (198)
=> R1 = 37.4169
Substitute R1 in (c) to get
R2 = 47.8639
QED



102909, 09:54 AM

#21

Senior Member
Join Date: Jul 2006
Location: Middle of the road, NJ
Bikes:
Posts: 2,568
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 17 Post(s)

Quote:
Originally Posted by oakback
My uncle asked my other uncle how electricity works. He replied, "It's magic, man. Don't mess with it."

As long as you don't let the magic blue smoke out, every thing is fine.



102909, 10:02 AM

#22

TShirt Guy
Join Date: Jul 2008
Location: Lansdale, PA
Bikes: 2005 Fuji Team Issue, 2007 Fuji SL1
Posts: 464
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Originally Posted by rajarajan
So 12/R1  12/(R1+R2) = 0.18
which leads to 12R2 = 0.18 * R1 * (R1+R2)  equation (a)
So 12/R2  12/(R1+R2) = 0.11
which leads to 12R1 = 0.11 * R2 * (R1+R2)  equation (b)
As one can see (R1+R2) common in the two equations, divide (b) by (a) and solve to get
R2*R2 = (R1*R1 * 18 ) / 11
leading to R2 = R1 * sqrt (18/11)  equation (c)
Substitute (c) in (a) to get
R1 * (1+ sqrt (18/11)) = 1200 / sqrt (198)
=> R1 = 37.4169
Substitute R1 in (c) to get
R2 = 47.8639
QED

QED?
Really?
1200 / sqrt(198) has two results. How can you QED when you don't specify which result of the square root is to be used?
just bustin' yer chops



102909, 08:55 PM

#23

Senior Member
Join Date: Sep 2006
Bikes: Kona Cinder Cone, Sun EZ3 AX
Posts: 1,195
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Quote:
Solving for I (the common variable in each equation), I get:
(i) I = (12 V)/(R1+R2)
(ii) I = [(12 V)/R1]  0.18A
(iii) I = [(12 V)/R2]  0.11A

At this point you have 3 equations with 3 unknowns. When you went on, you tried to solve using only equations ii and iii. You need to keep using equation i in your solution.



103009, 12:55 AM

#24

Member
Join Date: Aug 2006
Bikes:
Posts: 26
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)

Try reformulating as 12V = (R1+R2)I = R1(I+0.18) = R2(I+.11). Hint: what is I^2?



Thread Tools 
Search this Thread 


Posting Rules

You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is Off



All times are GMT 6. The time now is 10:39 AM.
