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Math Puzzle from March 29, 2019
Can you solve a differential equation in electrical engineering?
Written on by Noble Mushtak

The following was a puzzle presented to Marshwood GT students on March 29, 2019. Have fun doing math!

In AP Calculus AB, you learn about separable differential equations, which can seem like a really abstract technique which is not very practical. However, today, differential equations like these can actually be a very useful tool in physics and electrical engineering!

In this riddle, we're going to look at an RC circuit, which is a circuit with a charged capacitor and a resistor in a series circuit:

(Image from Wikimedia)

In terms of $$Q_0$$, the initial charge on the capacitor, $$C$$, the capacitance of the capacitor, and $$R$$, the resistance of the resistor, find an expression for $$Q(t)$$, the charge on the capacitor as a function of time.

Part 0: Most of you probably understand what a resistor is from high school physics, but some of you may have no idea what a capacitor is, which is important to solving this riddle. One simple way to think of a capacitor is as an electrical component which stores electrical potential energy, or voltage. A capacitor stores electrical potential energy by storing charge on two opposite plates, as shown in the picture below:

Here, the separation of the positive charge on the top plate and the negative charge on the bottom plate leads to an electric field which stores electrical potential energy. However, how do we calculate the voltage $$V$$ stored in the capacitor from the charge $$Q$$ on the capacitor's plate? It turns out that the voltage is actually directly proportional to the charge, and the constant of proportionality is called the capacitance, $$C$$, of the capacitor:

$$C=\frac{Q}{V}$$

Now, solve for the potential difference across the capacitor.

Part 1: Now that we've covered capacitors, let's move onto resistors. In high school physics, many of you learned about Ohm's Law, which says that the current flowing through the resistor is:

$$I=\frac{V}{R}$$

where $$V$$ is the voltage potential drop across the resistor and $$R$$ is the resistance of the resistor.

Now, in the RC circuit, let's think about where this current is coming from. Since all we have is a resistor and a capacitor, any current that flows through the resistor must also flow through the capacitor since current flows through the whole circuit. However, if current flows from the positively-charged plate to the negatively-charged plate of the capacitor, then the positively-charged plate must be losing charge, or becoming less positive, while the negatively-charged plate is gaining charge, or becoming less negative. This ultimately means that the current flowing through the resistor is the rate of decrease in $$Q$$, the charge in the capacitor, so we find that:

$$-\frac{dQ}{dt}=\frac{V}{R}$$

Now, solve for the voltage potential drop across the resistor.

Part 2: Kirchoff's loop rule says that around any loop in an electrical circuit, the sum of voltage sources, or EMFs, equals the sum of voltage potential drops. Thus, using your answers from Part 0 and Part 1, write down the differential equation that you find from Kirchoff's loop rule. Then, solve the differential equation for $$Q$$ as a function of time $$t$$, using the initial condition $$Q(0)=Q_0$$.

Click here to check your answer.The differential equation is $$\frac{Q}{C}=-R\frac{dQ}{dt}$$ and the solution is $$Q=Q_0e^{-\frac{t}{RC}}$$.

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