Chapter 2.2, Problem 11E

Interpretation Introduction

Interpretation:

Circuit solution can be obtained by solving differential equation analytically. The circuit solution for equation $\dot{\text{Q}}\text{=}\frac{{\text{V}}_{\text{0}}}{\text{R}}\text{-}\frac{\text{Q}}{\text{RC}}$ can be obtained by using the analytical solution of the given equation. Where the initial condition is $\text{Q(0)=0}$.

Concept Introduction:

The electrical circuit having a resistor $\text{R}$ and capacitor $\text{C}$ are in series with a battery of constant DC voltage ${\text{V}}_{\text{0}}$.

Suppose that the switch is closed at $\text{t=0}$, that means no charge on the capacitor initially.

As we go around the circuit, the total voltage drop always equal to zero. ${\text{V}}_0+RI+\frac{Q}{C}=0$

Where RI is the voltage across the resistor, I will be the current flowing through the resistor.

This current causes the charge to accumulate on the capacitor at a rate $\dot{Q}=I$.

By rearranging we get $\dot{\text{Q}}\text{=}\frac{{\text{V}}_{\text{0}}}{\text{R}}\text{-}\frac{\text{Q}}{\text{RC}}$.