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Jason Charney
Jason Charney

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Multiple Resistors in Parallel

In my previous post, I shared how to solve an parallel resistor equation with two resistors in parallel.

1RΣ2=1R1+1R2    RΣ2=R1R2R1+R2 \begin{aligned} \frac{1}{R_{\Sigma 2}} &= \frac{1}{R_1} + \frac{1}{R_2} &\iff R_{\Sigma 2} &= \frac{R_1 R_2}{R_1 + R_2} \end{aligned}

What if we had more than two resistors in parallel?

First, let's make a quick note about finding the resistance of circuits in series.

The total resistance of resistors in a series is simply the sum of their resistances.

RΣn=i=1n=R1+R2+R3++Rn \begin{aligned} R_{\Sigma n} = \sum_{i=1}^{n} = R_1 + R_2 + R_3 + \cdots + R_n \end{aligned}

So as long a the resistors are part of a path, they can be seen as one big resistor.

I could also probably go over Ohm's Law ( R=V/IR = V / I or V=IRV = IR where VV is voltage in volts, II is the electrical current in amperes, and RR is the resistance in ohms), but I don't want to do that. Maybe in another article.

This article is about parallel resistors.

Suppose we have three resistors in parallel.

1RΣ3=1R1+1R2+1R3 \begin{aligned} \frac{1}{R_{\Sigma 3}} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \end{aligned}

We can use the formula for two resistors in parallel, but we would need to add a third resistor.

1RΣ3=1R1+1R2+1R31RΣ3=1RΣ2+1R3RΣ2=R1R2R1+R21RΣ3=R1+R2R1R2+1R31RΣ3=R3(R1+R2)R1R2R3+R1R2R1R2R31RΣ3=R1R2+R1R3+R2R3R1R2R3RΣ3=R1R2R3R1R2+R1R3+R2R3 \begin{aligned} \frac{1}{R_{\Sigma 3}} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} && \square \cr \frac{1}{R_{\Sigma 3}} &= \frac{1}{R_{\Sigma 2}} + \frac{1}{R_3} &&\because R_{\Sigma 2} = \frac{R_1 R_2}{R_1 + R_2} \cr \frac{1}{R_{\Sigma 3}} &= \frac{R_1 + R_2}{R_1 R_2} + \frac{1}{R_3} \cr \frac{1}{R_{\Sigma 3}} &= \frac{R_3 (R_1 + R_2)}{R_1 R_2 R_3} + \frac{R_1 R_2}{R_1 R_2 R_3} \cr \frac{1}{R_{\Sigma 3}} &= \frac{R_1 R_2 + R_1 R_3 + R_2 R_3}{R_1 R_2 R_3} && \square \cr R_{\Sigma 3} &= \frac{R_1 R_2 R_3}{R_1 R_2 + R_1 R_3 + R_2 R_3} && \blacksquare \end{aligned}

We now have a formula we can use to find the resistance of three resistors in parallel.

What if we have four resistors in parallel? We can use the formula from our three parallel resistor example and use it to find a formula for four parallel resistors.

1RΣ4=1R1+1R2+1R3+1R41RΣ4=1RΣ3+1R4RΣ3=R1R2R3R1R2+R1R3+R2R31RΣ4=R1R2+R1R3+R2R3R1R2R3+1R41RΣ4=R4(R1R2+R1R3+R2R3)R1R2R3R4+R1R2R3R1R2R3R41RΣ4=R1R2R3+R1R2R4+R1R3R4+R2R3R4R1R2R3R4RΣ4=R1R2R3R4R1R2R3+R1R2R4+R1R3R4+R2R3R4 \begin{aligned} \frac{1}{R_{\Sigma 4}} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} && \square \cr \frac{1}{R_{\Sigma 4}} &= \frac{1}{R_{\Sigma 3}} + \frac{1}{R_4} &&\because R_{\Sigma 3} = \frac{R_1 R_2 R_3}{R_1 R_2 + R_1 R_3 + R_2 R_3} \cr \frac{1}{R_{\Sigma 4}} &= \frac{R_1 R_2 + R_1 R_3 + R_2 R_3}{R_1 R_2 R_3} + \frac{1}{R_4} \cr \frac{1}{R_{\Sigma 4}} &= \frac{R_4 (R_1 R_2 + R_1 R_3 + R_2 R_3)}{R_1 R_2 R_3 R_4} + \frac{R_1 R_2 R_3}{R_1 R_2 R_3 R_4} \cr \frac{1}{R_{\Sigma 4}} &= \frac{R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4}{R_1 R_2 R_3 R_4} && \square \cr R_{\Sigma 4} &= \frac{R_1 R_2 R_3 R_4}{R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4} && \blacksquare \end{aligned}

Right now, you are probably think to yourself "Ugh! That's a lot of things I can't track" or "My head hurts from looking at this!"

This is why when formulas like this get this complicated I like to shift towards Sigma ( Σ\Sigma ) and Pi ( Π\Pi ) notations for for series of sums and products.

The basic concepts of these notations is to avoid having to write things over and over. Let's say we wanted to find the sum of numbers 1 through 4.

i=14i=1+2+3+4=10 \begin{aligned} \sum_{i=1}^4 i = 1 + 2 + 3 + 4 = 10 \end{aligned}

Factorial ( n!n! ) is just a prettier way to write the product of a series of numbers starting at 1.

4!=i=14i=1234=24 \begin{aligned} 4! = \prod_{i=1}^4 i = 1 \cdot 2 \cdot 3 \cdot 4 = 24 \end{aligned}

But we can use Σ\Sigma and Π\Pi at different starting and ending points.

i=263i=6+9+12+15+18=60i=342+i=5679=1890 \begin{aligned} \sum_{i=2}^6 3i &= 6 + 9 + 12 + 15 + 18 = 60 \cr \prod_{i=3}^4 2+i &= 5 \cdot 6 \cdot 7 \cdot 9 = 1890 \cr \end{aligned}

Take a note of the formulas we have created thus far.

1RΣ2=1R1+1R2    RΣ2=R1R2R1+R21RΣ3=1R1+1R2+1R3    RΣ3=R1R2R3R1R2+R1R3+R2R31RΣ4=1R1+1R2+1R3+1R4    RΣ4=R1R2R3R4R1R2R3+R1R2R4+R1R3R4+R2R3R4 \begin{aligned} \frac{1}{R_{\Sigma 2}} &= \frac{1}{R_1} + \frac{1}{R_2} &\iff R_{\Sigma 2} &= \frac{R_1 R_2}{R_1 + R_2} \cr \frac{1}{R_{\Sigma 3}} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} &\iff R_{\Sigma 3} &= \frac{R_1 R_2 R_3}{R_1 R_2 + R_1 R_3 + R_2 R_3} \cr \frac{1}{R_{\Sigma 4}} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} &\iff R_{\Sigma 4} &= \frac{R_1 R_2 R_3 R_4}{R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4} \end{aligned}

Did you notice in the formulas on the right side, the numerators are the product of all the resistors?

If we could create a generalized formula, we could probably write the numerator in Π\Pi notation.

u=i=1nRi=R1R2R3Rn \begin{aligned} u &= \prod_{i = 1}^n R_i = R_1 \cdot R_2 \cdot R_3 \cdot \cdots \cdot R_n && \square \end{aligned}

Pretty easy to see, right?

The denominators will require a bit more understanding. Each item in the denominator is the numerator divided by one term to cancel out one of the factors in a term.

vj=uRj=i=1nRiRj=R1R2RjRn1RnRjwhere 1jn \begin{aligned} v_j &= \frac{u}{R_j} \cr &= \frac{\prod_{i = 1}^n R_i}{R_j} \cr &= \frac{R_1 \cdot R_2 \cdot \cdots \cdot \cancel{R_j} \cdot \cdots \cdot R_{n-1} \cdot R_n}{\cancel{R_j}} && \text{where } 1 \le j \le n \end{aligned}

But writing this is problematic. Nobody wants to write all that stuff I just wrote when they can use a Π\Pi notation to indicate that when we multiply by all our factors, we can skip RjR_j . That's where this neat little hack comes in.

vj=jiRj \begin{aligned} v_j &= \prod_{j \ne i} R_j \end{aligned}

Basically we say here "skip jj if it is equal to ii because jj should not equal ii ".

This is important because keep in mind, our denominator is the sum of products. Microsoft Excel geeks might now about the SUMPRODUCT function which acts as a filter add only matching items. With ji\prod_{j \ne i} , we basically added a filter to exclude multiplying RjR_j when j=ij = i .

There is a reason I put that subscripted jj next to vv . Its because we need the denominator ( vv ) to be the sum of each vjv_j of which each vjv_j is the product of all the resistors ( uu ) except when j=ij = i .

v=i=1nvi=v1+v2++vn=j1Rj+j2Rj++jnRjv=i=1nj1Rj \begin{aligned} v &= \sum_{i = 1}^n v_i \cr &= v_1 + v_2 + \cdots + v_n &= \prod_{j \ne 1} R_j + \prod_{j \ne 2} R_j + \cdots + \prod_{j \ne n} R_j \cr v &= \sum_{i = 1}^n \prod_{j \ne 1} R_j && \square \end{aligned}

Confused? Think of this as having two for loops. In the following code sample, our initial sum is zero and our initial product is one. Starting with a sum of one would throw our answer off by one, and starting with a product of zero would mean all the products are zero. We can't have that. I've opted to shift our product index by one because arrays in JavaScript start a zero, but our math examples start at one.

let r = {1.5, 1.5, 4, 4};  // each resister with values in ohms

function denominator(r){
    let sum = 0;
    for(let i = 1; i <= n; i++){     // sigma sum
        let product = 1;
        for(let j = 1; j <= n; j++){  // pi product
            if(i === j) continue;    // if i equals j, skip this step
            product *= r[j-1];
        sum += product;
    return sum;

console.log(denominator(r));   // => 66
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There's probably a more modern way to do this with JavaScript with the Array class's filter() or reduce() functions, but we still have to include our numerator in the formula!

For those wonder, here's what the code would be for the numerator.

function numerator(r) {
    let product = 1;
    for(i = 1; i <= n; i++){
        product *= r[i-1];
    return product;

console.log(numerator(r));   // => 36
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Let's combine both of our formulas now that we have two formulas to find each half of this fraction.

RΣn=i=1nRii=1nj1Rj \begin{aligned} R_{\Sigma n} &= \frac{\prod_{i = 1}^n R_i}{\sum_{i = 1}^n \prod_{j \ne 1} R_j} &&\blacksquare \end{aligned}

One more piece of code.

function parallel(r){
    return numerator(r)/denominator(r);

console.log(numerator(r));   // => 0.54545454545
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But I didn't spend about a hour writing thsi just to show you this.

Remember last post where we wanted to find the rates of change in a resistor? There's a reason I used uu and vv in this example, because we're not done yet! We want to find the derivative of RΣnR_{\Sigma n} .

Get the ibuprofen handy, because I'm going to make your head hurt.

Because RΣnR_{\Sigma n} is the quotient of two values, so will our RΣnR_{\Sigma n}' . Which means, we will need to break out the the quotient rule.

RΣn(t)=(uv)=uvuv[v]2 \begin{aligned} R_{\Sigma n}'(t) &= \left(\frac{u}{v}\right)' = \frac{u' \cdot v - u \cdot v'}{[v]^2} \end{aligned}

But I'm getting ahead of myself. I mean, how do we derive a Σ\Sigma series, or a Π\Pi series, or a ΣΠ\Sigma\Pi series!?

We should probably start by taking a look at how we derived a circuit with two parallel resistors in it, from the previous post.

RΣ2(t)=R1(t)R2(t)R1(t)+R2(t)    RΣ2(t)=R1(t)[R2(t)]2+[R1(t)]2R2(t)[R1(t)+R2(t)]2 \begin{aligned} R_{\Sigma 2}(t) &= \frac{R_1(t) R_2(t)}{R_1(t) + R_2(t)} &\implies R_{\Sigma 2}'(t) &= \frac{R_1'(t) \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot R_2'(t)}{[R_1(t) + R_2(t)]^2} \end{aligned}

You know what, let's drop the (t)(t) 's. I know they are important to remind ourselves that we are working with functions with respect to time, and there are many ways to write your notation. But Team Lagrange for the win.

RΣ2=R1R2R1+R2    RΣ2=R1[R2]2+[R1]2R2[R1+R2]2 \begin{aligned} R_{\Sigma 2} &= \frac{R_1 R_2}{R_1 + R_2} &\implies R_{\Sigma 2}' &= \frac{R_1' \cdot [R_2]^2 + [R_1]^2 \cdot R_2'}{[R_1 + R_2]^2} \end{aligned}

OK, so that was our formula for two resistors in parallel. What about three? The numerator is easy to figure out. The product rule indicates that one of the values in each term is the first derivative. The denominator will require the sum rule and the product rule.

For brevity, I decided not to include some of my work because there was a lot of F.O.I.L. and cancelling. You can expect the same when I work on four factors in a bit. The point here is to find patterns.

RΣ3=R1R2R3R1R2+R1R3+R2R3RΣ3=uvRΣ3=uvuv[v]2u=R1R2R3u=(R1R2R3)u=R1R2R3+R1R2R3+R1R2R3v=R1R2+R1R3+R2R3v=(R1R2+R1R3+R2R3)v=(R1R2)+(R1R3)+(R2R3)v=R1R2+R1R2+R1R3+R1R3+R2R3+R2R3v2=[R1R2+R1R3+R2R3]2uvuv=R1[R2]2[R3]2+[R1]2R2[R3]2+[R1]2[R2]2R3RΣ3=R1[R2]2[R3]2+[R1]2R2[R3]2+[R1]2[R2]2R3[R1R2+R1R3+R2R3]2 \begin{aligned} R_{\Sigma 3} &= \frac{R_1 R_2 R_3}{R_1 R_2 + R_1 R_3 + R_2 R_3} \cr R_{\Sigma 3} &= \frac{u}{v} \cr R_{\Sigma 3}' &= \frac{u' \cdot v - u \cdot v'}{[v]^2}\cr \cr u &= R_1 R_2 R_3 \cr u' &= (R_1 R_2 R_3)' \cr u' &= R_1' R_2 R_3 + R_1 R_2' R_3 + R_1 R_2 R_3' && \square \cr \cr v &= R_1 R_2 + R_1 R_3 + R_2 R_3 \cr v' &= (R_1 R_2 + R_1 R_3 + R_2 R_3)' \cr v' &= (R_1 R_2)' + (R_1 R_3)' + (R_2 R_3)' \cr v' &= R_1' R_2 + R_1 R_2' + R_1' R_3 + R_1 R_3' + R_2' R_3 + R_2 R_3' && \square \cr \cr v^2 &= [R_1 R_2 + R_1 R_3 + R_2 R_3]^2 &&\square \cr \cr u' \cdot v - u \cdot v' &= R_1' [R_2]^2 [R_3]^2 + [R_1]^2 R_2' [R_3]^2 + [R_1]^2 [R_2]^2 R_3' && \square \cr R_{\Sigma 3}' &= \frac{R_1' [R_2]^2 [R_3]^2 + [R_1]^2 R_2' [R_3]^2 + [R_1]^2 [R_2]^2 R_3'}{[R_1 R_2 + R_1 R_3 + R_2 R_3]^2} && \blacksquare \end{aligned}

That took me over an hour to write, not including the stuff I cut out for clarity.

So what can we learn from this? Firstly, we don't need to expand the denominator. Just don't. It's a lot of busy work. In fact, I won't be expanding the denominator in or four parallel resistor example or in the nn th example. So lets assume that in any first derivative, of RΣnR_{\Sigma n} , the denominator is as follows

[vΣn]2=[i=1nj1Rj]2 \begin{aligned} [v_{\Sigma n}]^2 &= \left[\sum_{i = 1}^n \prod_{j \ne 1} R_j\right]^2 && \square \end{aligned}

Our attention is primarily on the numerator.

We can see that the terms in the numerator for RΣ2R_{\Sigma 2} are multiplied by [R3]2[R_{3}]^2 . If we group our numerator terms together, we see that each numerator term is a denominator term multiplied by the derivative of a resistor. We will need to prove this in our four parallel resistor example to validate our generalization.

1RΣ4=1R1+1R2+1R3+1R4RΣ4=R1R2R3R4R1R2R3+R1R2R4+R1R3R4+R2R3R4RΣ4=(R1R2R3R4)(R1R2R3+R1R2R4+R1R3R4+R2R3R4)u=R1R2R3R4u=(R1R2R3R4)=R1R2R3R4R1R2R3R4R1R2R3R4R1R2R3R4v=R1R2R3+R1R2R4+R1R3R4+R2R3R4v=(R1R2R3+R1R2R4+R1R3R4+R2R3R4)=(R1R2R3)+(R1R2R4)+(R1R3R4)+(R2R3R4)=R1R2R3+R1R2R3+R1R2R3+R1R2R4+R1R2R4+R1R2R4+R1R3R4+R1R3R4+R1R3R4+R2R3R4+R2R3R4+R2R3R4v2=[R1R2R3+R1R2R4+R1R3R4+R2R3R4]2uvuv=R1[R2]2[R3]2[R4]2+[R1]2R2[R3]2[R4]2+[R1]2[R2]2R3[R4]2+[R1]2[R2]2[R3]2R4RΣ4=R1[R2]2[R3]2[R4]2+[R1]2R2[R3]2[R4]2+[R1]2[R2]2R3[R4]2+[R1]2[R2]2[R3]2R4[R1R2R3+R1R2R4+R1R3R4+R2R3R4]2 \begin{aligned} \frac{1}{R_{\Sigma 4}} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \cr R_{\Sigma 4} &= \frac{R_1 R_2 R_3 R_4}{R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4} \cr \cr R_{\Sigma 4}' &= \frac{(R_1 R_2 R_3 R_4)'}{(R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4)'} \cr \cr u &= R_1 R_2 R_3 R_4 \cr u' &= (R_1 R_2 R_3 R_4)' \cr &= R_1' R_2 R_3 R_4 R_1 R_2' R_3 R_4 R_1 R_2 R_3' R_4 R_1 R_2 R_3 R_4' &&\square\cr \cr v &= R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4 \cr v' &= (R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4)' \cr &= (R_1 R_2 R_3)' + (R_1 R_2 R_4)' + (R_1 R_3 R_4)' + (R_2 R_3 R_4)' \cr &= R_1' R_2 R_3 + R_1 R_2' R_3 + R_1 R_2 R_3' \cr &+ R_1' R_2 R_4 + R_1 R_2' R_4 + R_1 R_2 R_4' \cr &+ R_1' R_3 R_4 + R_1 R_3' R_4 + R_1 R_3 R_4' \cr &+ R_2' R_3 R_4 + R_2 R_3' R_4 + R_2 R_3 R_4' \cr \cr v^2 &= [R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4]^2 \cr u'v - uv' &= R_1' [R_2]^2 [R_3]^2 [R_4]^2 \cr &+ [R_1]^2 R_2' [R_3]^2 [R_4]^2 \cr &+ [R_1]^2 [R_2]^2 R_3' [R_4]^2 \cr &+ [R_1]^2 [R_2]^2 [R_3]^2 R_4' \cr \cr R_{\Sigma 4}' &= \frac{ R_1' [R_2]^2 [R_3]^2 [R_4]^2 + [R_1]^2 R_2' [R_3]^2 [R_4]^2 + [R_1]^2 [R_2]^2 R_3' [R_4]^2 + [R_1]^2 [R_2]^2 [R_3]^2 R_4' }{[R_1 R_2 R_3 + R_1 R_2 R_4 + R_1 R_3 R_4 + R_2 R_3 R_4]^2} && \blacksquare \end{aligned}

See how complex this is. Fortunately, we can see our pattern.

With every new factor, for each product term, at least one of the factors is derived while the others are squared. With this information, we can generalize the numerator. Let's start with how to describe each term

[ui]=Riji[Rj]2 \begin{aligned} [u_i]' &= R_i' \prod_{j\ne i} [R_j]^2 \end{aligned}

Each subterm is the product of the derivative multiplied by the squares of the other factors except when j=ij = i

We can also describe the sum of these subterms. This will describe our numerator.

uΣn=i=1n(Riji[Rj]2) \begin{aligned} u_{\Sigma n}' &= \sum_{i=1}^n \left( R_i' \prod_{j\ne i} [R_j]^2 \right) &&\square \end{aligned}

We have reached the end of a long journey. All there is now is to pair our numerator with the denominator.

RΣn=i=1n(Riji[Rj]2)[i=1nj1Rj]2 \begin{aligned} R_{\Sigma n}' &= \frac{\sum_{i=1}^n \left( R_i' \prod_{j\ne i} [R_j]^2 \right)}{ \left[ \sum_{i = 1}^n \prod_{j \ne 1} R_j \right]^2 } && \blacksquare \end{aligned}

Let's compare this derivative with the formula

1RΣn(t)=i=1n1Ri(t)    RΣn(t)=i=1nRi(t)i=1nj1Rj(t)    RΣn(t)=i=1n(Ri(t)ji[Rj(t)]2)[i=1nj1Rj(t)]2 \begin{aligned} \frac{1}{R_{\Sigma n}(t)} &= \sum_{i=1}^n \frac{1}{R_i(t)} &\iff R_{\Sigma n}(t) &= \frac{\prod_{i=1}^n R_i(t)}{\sum_{i = 1}^n \prod_{j \ne 1} R_j(t)} \cr &&\implies R_{\Sigma n}'(t) &= \frac{\sum_{i=1}^n \left( R_i'(t) \prod_{j\ne i} [R_j(t)]^2 \right)}{ \left[ \sum_{i = 1}^n \prod_{j \ne 1} R_j(t) \right]^2 } &&\blacksquare \end{aligned}

I wanted to interpret the derived formula into code, but I just don't have the energy left after having figure out all the terms for four resistors in parallel. Maybe in a future post. Stay tuned.

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