Missing resistance in parallel circuit

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mwootan

Joined Apr 16, 2009
15
I am a bit confused on figuring out a circuit. This is a parallel circuit with 3 resisters.
R1 = ?
R2 = 100Ω
R3 = 50Ω
120 volts to circuit
It = 9.6A
How do I figure out what the value of R1 is?
Hope someone can point me in the right direction??? Shmurk

Joined Aug 13, 2007
24
If It is the total current entering the branching (using Ohm's law, or Kirchhoff's current law)

- compute I2, and I3.
- compute I1 with I2, I3, and It.
- compute R1.
- finished!

Last edited:

mwootan

Joined Apr 16, 2009
15
If I have resistance resistance R2 and R3 and need R1 is this correct to find missing R1?
R2 = 100 Ω
R3 = 50 Ω
120 Vac
It = 9.6A
I figured out Ir2 and Ir3
Ir2 = 1.2A
Ir3 = 2.4A

I take It = 9.6 and subtract Ir2 and Ir3 which leaves me Ir1 = 6A
Using R = E/I R= 120v/6A so R1 = 20 Ω
Did I figure this correctly? SgtWookie

Joined Jul 17, 2007
22,201
Yes. Good job. Everett S.

Joined Aug 5, 2014
1
First, you have to follow ohm's law to find the total resistance.

120/9.6 = 12.5 Ω.(this is your Rt)

next, this gets a little more complicated, the Rt = the reciprocal (1/) of the sum of reciprocal of the individual resistors (1/R1 + 1/R2 + 1/R3.)
The easiest way to break this down is to to lay it out algebraically, so lets do it!

12.5 = 1/
1/100 + 1/50 + 1/R3

Now that we have an algebraic look at it, we can can use it to solve for the missing resistor value.

1) Apply the inverse function to both sides of the equation. This will cancel out the first reciprocal, and will manipulate the equation as so.

12.5 = 1/
1/100 + 1/50 + 1/R3 (use the inverse function (1/) on both sides. this cancels out the first reciprocal on the right side of the equation, the 1/ over the values is gone and 1/Rt appears)

1/12.5 = 1/100 + 1/50 + 1/R3 (afterwards)

2) Apply the reciprocals to each term (actually divide 1 by each term) to obtain the appropriate values. This will manipulate the equation as so.

1/12.5 = 1/100 + 1/50 + 1/R3

.08 = .01 + .02 + R3 (afterwards)

3) Now that we have the actual values, combine like terms. This will manipulate the equation as so.

.08 = (.01 + .02) + R3

.08 = .03 + R3
-.03 -.03

.05 = R3

4) now that we have the reciprocal value of the missing resistor, apply the inverse function. this should give us the original value of the missing resistor.

1/.05 = 20Ω (the missing resistor value)

Hope you and others can use this as a blueprint for future problems.

djsfantasi

Joined Apr 11, 2010
5,473
@Everett. Homework Help is just that; not Homework Done for You. Typically, when responding to posts in this sub-forum, it's appropriate to lead the OP to the solution and not appropriate to provide one.

Having said that, in your step 2, I doubt your math that turns 1/R3 into R3. When taking the inverse, it applies to the entire right side of the equation, not individual terms.

MrChips

Joined Oct 2, 2009
19,157
@dj - You are correct in saying you cannot invert the single terms in an inversion. OP is not performing an inversion. OP is evaluating each term as is.

@ Everett - While your method is sound, your mathematical notation is not. When you perform the reciprocal you have to retain the 1/R3.

.08 = .01 + .02 + R3 (afterwards)

should be written as:

.08 = .01 + .02 + 1/R3

While you arrive at the correct result, evaluating the individual currents and subtracting from the total as shown by the OP in post #4 still appears to me to be a simpler solution.

djsfantasi

Joined Apr 11, 2010
5,473
MrChips,

I saw that he was evaluating the terms, but was confused when it was implied that R3=1/R3

crutschow

Joined Mar 14, 2008
23,144
.....................
...................... evaluating the individual currents and subtracting from the total as shown by the OP in post #4 still appears to me to be a simpler solution.
I agree. There's usually more than one way to solve for unknowns in a circuit, but the simplest is usually the best. Easier to do and less likely to generate an error in the calculations. And I try to avoid any solutions that involve solving simultaneous equations. t_n_k

Joined Mar 6, 2009
5,448
Given this problem was first posted on 16th April 2009 I doubt the OP is still puzzling over the solution....

MrChips

Joined Oct 2, 2009
19,157
I didn't even notice.

Tony Montoya

Joined Sep 22, 2015
1
Formula and Easy way on calculator 1÷((1÷rt)-(1÷R1)-(1÷r3))=
Example
Rt=3.582,
r1=16
r2=10
r3=?
r4=20
If You need to find the value for r3
1÷((1÷3.582)-(1÷16)-(1÷10)-(1÷20)) = 15
R3 = 15 ohms

WBahn

Joined Mar 31, 2012
24,573
You've just practiced the arcane art of necroposting -- reviving a long-dead thread. Since it has been more than a year since this thread was active, it is unlikely that the TS has any need or desire for further discussion.

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