Resistors in Parallel
Tags: mathematics, derivatives, calculus, resistors
Published: 2021-10-20 17:03:04

---
title: Resistors in Parallel
published: true
description: How to solve for resistors in parallel
tags: mathematics, derivatives, calculus, resistors
cover_image: https://dev-to-uploads.s3.amazonaws.com/uploads/articles/rnb7284lk12ow3woatp5.png 
---

Hello, this is my first post here. One of the things I like to do in my spare time is generalize mathematical formulas.

On the subreddit [r/calculus](https://www.reddit.com/r/calculus), [this question](https://www.reddit.com/r/calculus/comments/qbgrgi/related_rates_i_dont_understand_what_the_question/) was posted.

![Question](https://dev-to-uploads.s3.amazonaws.com/uploads/articles/ikgjdd84fl5eex1hz13f.png)

The question essentially asks what is the change in resistance with respect to time.

The formula for two resistors in parallel is the following.

{% katex %}
\frac{1}{R_{\Sigma 2}} = \frac{1}{R_1} + \frac{1}{R_2}
{% endkatex %}

> Note: I used the {% katex inline %}R_{\Sigma 2}{% endkatex %} to indicate the Sum of Two Resistors. ({% katex inline %}\Sigma{% endkatex %} often used for "Sum". I will use this notation in future posts.

What trips up a lot of students with this problem is that they feel overwhelmed by how they should calculate the problem.

To fix that, they should adjust the formula before plug in the numbers.

{% katex %}
\begin{aligned}
\frac{1}{R_{\Sigma 2}} &= \frac{1}{R_1} + \frac{1}{R_2} && \square \cr
\frac{1}{R_{\Sigma 2}} &= \frac{R_2}{R_1 R_2} + \frac{R_1}{R_1 R_2} \cr
\frac{1}{R_{\Sigma 2}} &= \frac{R_1}{R_1 R_2} + \frac{R_2}{R_1 R_2} \cr
\frac{1}{R_{\Sigma 2}} &= \frac{R_1 + R_2}{R_1 R_2} && \square \cr
R_{\Sigma 2} &= \frac{R_1 R_2}{R_1 + R_2} && \blacksquare
\end{aligned}
{% endkatex %}

Since this is a problem with respect to time, we need to think of this problem like this.

{% katex %}
\frac{1}{R_{\Sigma 2}(t)} = \frac{1}{R_1(t)} + \frac{1}{R_2(t)}
{% endkatex %}

Or, since we adjusted our formula, this.

{% katex %}
R_{\Sigma 2}(t) = \frac{R_1(t) \cdot R_2(t)}{R_1(t) + R_2(t)}
{% endkatex %}

We need to find the derivative of this problem.

{% katex %}
R'_{\Sigma 2}(t) = \left( \frac{R_1(t) \cdot R_2(t)}{R_1(t) + R_2(t)} \right)'
{% endkatex %}

Let's make some substitutions. Let us set the numerator to {% katex inline %}u(t){% endkatex %} and the denominator to {% katex inline %}v(t){% endkatex %}.

{% katex %}
R'_{\Sigma 2}(t) = \left(\frac{u(t)}{v(t)}\right)' = \left( \frac{R_1(t) \cdot R_2(t)}{R_1(t) + R_2(t)} \right)'
{% endkatex %}

We can apply the [**quotient rule**](https://en.wikipedia.org/wiki/Quotient_rule) to our equation.

{% katex %}
R'_{\Sigma 2}(t) = \left(\frac{u(t)}{v(t)}\right)' = \frac{u'(t) \cdot v(t) - u(t) \cdot v'(t)}{[v(t)]^2}
{% endkatex %}

Let us find the derivatives of {% katex inline %}u(t){% endkatex %} and {% katex inline %}v(t){% endkatex %}.  We will apply the [**product rule**](https://en.wikipedia.org/wiki/Product_rule) to {% katex inline %}u'(t){% endkatex %} and the [**Sum Rule**](https://en.wikipedia.org/wiki/Differentiation_rules#Differentiation_is_linear) to {% katex inline %}v'(t){% endkatex %}.

{% katex %}
\begin{aligned}
u(t) &= R_1(t) \cdot R_2(t) &\implies u'(t) &= R_1'(t) \cdot R_2(t) + R_1(t) \cdot R_2'(t) && \square \cr
v(t) &= R_1(t) + R_2(t) &\implies v'(t) &= R_1'(t) + R_2'(t) && \square
\end{aligned}
{% endkatex %}

Since the denominator of the quotient rule is {% katex inline %}[v(t)]^2{% endkatex %}, we should find the square of {% katex inline %}v(t){% endkatex %}.

{% katex %}
\begin{aligned}
[v(t)]^2 &= [R_1(t) + R_2(t)]^2 \cr
 &= [R_1(t)]^2 + 2 \cdot R_1(t) \cdot R_2(t) + [R_2(t)]^2 && \square
\end{aligned}
{% endkatex %}

Let's return to the numerator, {% katex inline %}u'(t) \cdot v(t) - u(t) \cdot v'(t){% endkatex %}. We should find {% katex inline %}u'(t) \cdot v(t){% endkatex %} and {% katex inline %}u(t) \cdot v'(t){% endkatex %}. This is quite long, so let's wrap this around each term.

{% katex %}
\begin{aligned}
u'(t) \cdot v(t) &= (R_1'(t) \cdot R_2(t) + R_1(t) \cdot R_2'(t)) \cdot (R_1(t) + R_2(t)) \cr
&= R_1'(t) \cdot R_2(t) \cdot R_1(t) \cr
&+ R_1'(t) \cdot R_2(t) \cdot R_2(t) \cr
&+ R_1(t) \cdot R_2'(t) \cdot R_1(t) \cr
&+ R_1(t) \cdot R_2'(t) \cdot R_2(t) \cr
\cr
u(t) \cdot v'(t) &= (R_1(t) \cdot R_2(t)) \cdot (R_1'(t) + R_2'(t)) \cr
&= R_1(t) \cdot R_2(t) \cdot R_1'(t) \cr
&+ R_1(t) \cdot R_2(t) \cdot R_2'(t)
\end{aligned}
{% endkatex %}

Recall that there is a minus sign in front of {% katex inline %}u(t) \cdot v'(t){% endkatex %}. We will need to negate the sign on the two terms distributed. We will also group similar terms in the {% katex inline %}u'(t) \cdot v(t){% endkatex %} equation. We will remove the terms that cancel each other out, and add what remains to get our numerator.

{% katex %}
\begin{aligned}
u'(t) \cdot v(t) - u(t) \cdot v'(t) 
&= (R_1'(t) \cdot R_2(t) + R_1(t) \cdot R_2'(t)) 
\cdot (R_1(t) + R_2(t)) \cr
&= \cancel{R_1'(t) \cdot R_2(t) \cdot R_1(t)} \cr
&+ R_1'(t) \cdot [R_2(t)]^2 \cr
&+ [R_1(t)]^2 \cdot R_2'(t) \cr
&+ \bcancel{R_1(t) \cdot R_2'(t) \cdot R_2(t)} \cr
&- \cancel{R_1(t) \cdot R_2(t) \cdot R_1'(t)} \cr
&- \bcancel{R_1(t) \cdot R_2(t) \cdot R_2'(t)} \cr
&= R_1'(t) \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot R_2'(t) && \square
\end{aligned}
{% endkatex %}

Let's put our equations together.

{% katex %}
\begin{aligned}
R_{\Sigma 2}'(t) &= \frac{R_1'(t) \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot R_2'(t)}{[R_1(t)]^2 + 2 \cdot R_1(t) \cdot R_2(t) + [R_2(t)]^2} && \square
\end{aligned}
{% endkatex %}

You know what, let's factor the denominator. It seems pointless to expand it.

{% katex %}
\begin{aligned}
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} && \blacksquare
\end{aligned}
{% endkatex %}

Perfect!

Take note that in the numerator there is no negative sign like is typical in a quotient rule. Because we cancelled out some terms, we can simply add everything in the numerator.

We can now solve the problem with that equation.

Let's consider {% katex inline %}R_1'(t) = 1\Omega/\text{s}{% endkatex %} and {% katex inline %}R_2'(t) = 1.5\Omega/\text{s}{% endkatex %}. (These are our derived values because these values are in ohms per second, so resistance with respect to time.)

If {% katex inline %}R_1(t) = 50\Omega{% endkatex %} and {% katex inline %}R_2(t) = 75\Omega{% endkatex %}, what is the rate of change of {% katex inline %}R(t){% endkatex %}?

{% katex %}
\begin{aligned}
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} \cr
R_{\Sigma 2}'(t) &= \frac{1 \cdot [R_2(t)]^2 + [R_1(t)]^2 \cdot 1.5}{[R_1(t) + R_2(t)]^2} && R_1'(t) = 1\Omega/\text{s}, R_2'(t) = 1.5\Omega/\text{s} \cr
R_{\Sigma 2}'(t) &= \frac{1 \cdot [75]^2 + [50]^2 \cdot 1.5}{[50 + 75]^2} && R_1(t) = 50\Omega, R_2(t) = 75\Omega\cr
R_{\Sigma 2}'(t) &= \frac{1 \cdot 5625 + 2500 \cdot 1.5}{[125]^2}\cr
R_{\Sigma 2}'(t) &= \frac{5625 + 3750}{15625}\cr
R_{\Sigma 2}'(t) &= \frac{9375}{15625}\cr
R_{\Sigma 2}'(t) &= 0.6\Omega/\text{s} && \blacksquare
\end{aligned}
{% endkatex %}

Our solution is **0.6 ohms per second**. 😎

This was a wonderful problem to solve.

In my next post, I want to show how we could generalize the formula we used.  What if we had more than two resistors in parallel.

I look forward to using this blog to share math problems as well as code examples.  Thanks for reading!