# How to make total resistance of 1 ohm?

• Abs321
In summary, the equation for parallel circuit is: 1/R1+1/R2=1/R. The equation for serial circuit is: R1+R2=R. The equation for a circuit with one resistor is: R=R1+R2. The equation for a circuit with two resistors is: R1+R2=1/R. The equation for a circuit with three resistors is: R1+R2=1/R1+1/R2. The equation for a circuit with four resistors is: R1+R2=1/R1+1/R2+1/R3. The equation for a circuit with five resistors is:
Abs321

## Homework Statement

We have 1Ω, 2Ω, 2Ω, 4Ω, 5Ω and 6Ω resistors. We have to make a circuit with all of them and total resistance must be 1 ohm.

## Homework Equations

Parallel: 1/R1+1/R2=1/R
In series: R1+R2=R

## The Attempt at a Solution

I have only made a circuit from 4 resistors:
1 ohm and 2 ohms in series, 6 ohms parallel to them and 2 ohms parallel to them all

Hi Abs,

Unfortunately, that doesn't count as an attempt (more like an arbitrary guess).
You need a bit more than that; considerations, for example. Like:
Your last step will be a parallel circuit (because a series will exceed 1 ##\Omega##).
That parallel circuit will have ##\ge## two sub-circuits, etc. etc.
Fortunately, I don't have the answer either. so you must take the first next step

 With a little trial and error I stumbled on the (or: a ) solution. How far are you ?

Last edited:
Oh, I suppose there could be two parallel circuit from 2 ohms and another sub-circuit totally of 2 ohms and so on, I understand how these laws work, but I can't construct such circuit. Just playing on with resistors.

Yes, one of the considerations is: the 1 ##\Omega## can't be a single parallel branch (total would go under 1 ##\Omega##).

So the first candidate as a parallel branch would be the 2 ##\Omega##. That reduces the problem to making another 2 ##\Omega## with the remaining 5 resistors (1,1,4,5,6) (since 2 ##\Omega\, \parallel \, 2 \Omega = 1 \Omega ## )

And so on. You're almost there !

And this is where I stop. Is it possible to combine 2 ohms in 2 parallel branches from remaining 5 resistors? Or do we need 3 branches?

Yes. So in total you end up with three branches: one of 2 ohm.
The remainder needs to be 2 ohm. Can't be one branch (I think -- why not ?) but two branches is worth investigating
 turns out to be mistaken. See below .

Last edited:
Oh, I've just found an answer. It's (6 ∥ 5+1∥ 4+2) ∥ 2

Brilliant ! Well done.

My solution was for the wrong problem (1,1,2,4,5,6) instead of (1,2,2,4,5,6). Want to try that too ?

Greg Bernhardt

## 1. How do I calculate total resistance of a circuit?

The total resistance of a circuit can be calculated by adding up the individual resistances in the circuit. This can be done using Ohm's Law (R = V/I) or by using the formula for resistors in series (R = R1 + R2 + R3...).

## 2. What is the unit of measurement for resistance?

The unit of measurement for resistance is the ohm (Ω), which is represented by the Greek letter omega.

## 3. Can I make a circuit with a total resistance of 1 ohm using only one resistor?

Yes, it is possible to make a circuit with a total resistance of 1 ohm using only one resistor. However, the resistor must have a resistance of 1 ohm or be adjustable to a resistance of 1 ohm.

## 4. Can I change the total resistance of a circuit?

Yes, the total resistance of a circuit can be changed by adding or removing resistors or by adjusting the resistance of existing resistors. Additionally, the length and thickness of the wires in the circuit can also affect the total resistance.

## 5. Why is it important to have a total resistance of 1 ohm in a circuit?

Having a total resistance of 1 ohm in a circuit allows for a predictable flow of current. It also allows for easier calculations when using Ohm's Law to determine voltage and current in the circuit. Additionally, it is a commonly used value for testing and calibrating electronic equipment and components.

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