Question

A series circuit contains a resistor and a capacitor as shown in Figure 1.3.15. Determine a differential equation for the charge q(t) on the capacitor if the resistance is R, the capacitance is C, and the impressed voltage is E(t).

Short answer

The differential equation is $R\frac{dq}{dt}+\frac{1}{C}q=E\left(t\right)$.

Step-by-step solution

Kirchhoff’s law:

It states that the sum of the voltages going across the resistor and the capacitor is equal to the voltage across the battery.

$V_{\text{resistor}}+V_{\text{capacitor}}=E\left(t\right).$

Apply Kirchhoff’s & Ohm’s law

Let q be the charge flows in the circuit, I be the current, C be the capacitance, and R be the resistance.

By Ohm's Law, the current passing through the resistor is V=IR and the voltage across the capacitor is V=q/C.

Using Kirchhoff’s law & ohm’s law we have:

$IR+\frac{q}{C}=E\left(t\right)$--------(1)

Evaluate

Using the definition of current i.e; current is defined as rate of flow of charge we have:

$I=\frac{dq}{dt}$

Substitutein $I=\frac{dq}{dt}$equation (1) to get:

$\frac{dq}{dt}R+\frac{q}{C}=E\left(t\right)$

Therefore, the differential equation for the total voltage drops across the entire circuit is:

$R\frac{dq}{dt}+\frac{1}{C}q=E\left(t\right)$