# Foo - Help with electric circuit problem!

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darksiderising
10-28-09, 12:42 PM
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:

Two resistances, R1 and R2, are connected in series across a 12-V 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.

KingTermite
10-28-09, 12:44 PM
resistors in parallel or series?

darksiderising
10-28-09, 12:46 PM
resistors in parallel or series?

Series, as stated in the problem.

KingTermite
10-28-09, 12:50 PM
Series, as stated in the problem.

Sorry...duh, I missed that. I wanted to make sure you were setting up the problems correctly.

KingTermite
10-28-09, 12:55 PM
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.

black_box
10-28-09, 01:01 PM
nevermind...

KingTermite
10-28-09, 01:02 PM
When you're ready, highlight:Don't do his homework for him...just help him get in the right direction.

black_box
10-28-09, 01:14 PM
i figured he'd have to show his work anyway?

ritepath
10-28-09, 01:30 PM
would have been more useful if they'd gave you the total current draw with R1 & R2 in circuit...

DannoXYZ
10-28-09, 01:55 PM
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.

darksiderising
10-28-09, 02:31 PM
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:

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.

rajarajan
10-28-09, 03:20 PM
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

darksiderising
10-28-09, 05:53 PM
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.

deraltekluge
10-28-09, 06:24 PM
Never mind...I'll leave it for rajarajan to explain.

sknhgy
10-28-09, 06:55 PM
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.

darksiderising
10-28-09, 06:59 PM
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.

sknhgy
10-28-09, 07:21 PM
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.

darksiderising
10-28-09, 07:55 PM
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.

black_box
10-28-09, 08:57 PM
(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.

rajarajan
10-29-09, 08:46 AM
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)

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

leob1
10-29-09, 09:54 AM
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.

ehidle
10-29-09, 10:02 AM
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?

:p

just bustin' yer chops :D

deraltekluge
10-29-09, 08:55 PM
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.

sisyphus321
10-30-09, 12:55 AM
Try reformulating as 12V = (R1+R2)I = R1(I+0.18) = R2(I+.11). Hint: what is I^2?