I am currently studying the mathematics textbook Nonlinear Dynamics and Chaos by Strogatz. As part of an example, I am provided with the following electrical circuit and accompanying explanation:

 

Example 2.2.2: Consider the electrical circuit shown in Figure 2.2.3. A resistor \$ R \$ and a capacitor \$ C \$ are in series with a battery of constant dc voltage \$ V_0 \$. Suppose that the switch is closed at \$ t = 0 \$, and that there is no charge on the capacitor initially. Let \$ Q(t) \$ denote the charge on the capacitor at time \$ t \ge 0 \$. Sketch the graph of \$ Q(t) \$.

 

Solution: This type of circuit problem is probably familiar to you. It is governed by linear equations and can be solved analytically, but we prefer to illustrate the geometric approach.

 

First we write the circuit equations. As we go around the circuit, the total voltage drop must equal zero; hence \$ -V_0 + RI + Q/C = 0 \$, where \$ I \$ is the current flowing through the resistor. This current causes charge to accumulate on the capacitor at a rate \$ \dot{Q} = I \$. Hence

 

$$ -V_0 + RQ + Q/C = 0 $$

 

or

 

$$\dot{Q} = f(Q) = \dfrac{V_0}{R} - \dfrac{Q}{RC}.$$

I have started studying electronics, but I'm still learning the basics, and haven't yet learned enough to understand the electronics reasoning for what the author has written. Specifically, I'm wondering about the following:

1. Kirchhoff's voltage law states that the directed sum of the voltages around a circuit equal zero, but what is the reason for the terms of the equation \$ -V_0 + RI + Q/C = 0 \$ being structured as they are? Specifically, why is the voltage given a negative sign, why is the resistance multiplied by the current, and why is the charge on the capacitor divided by the capacitance?

2. I understand that \$ -V_0 + RI + Q/C = 0 \Rightarrow \dot{Q} = f(Q) = \dfrac{V_0}{R} - \dfrac{Q}{RC} \$ by algebra, but why is this equal to \$ f(Q) \$? In other words, why is the rate at which charge accumulates on the capacitor, \$ \dot{Q} \$, a function of the charge of the capacitor, \$ f(Q) \$?

I would greatly appreciate it if people would please take the time to clarify these points.

I am currently studying the mathematics textbook Nonlinear Dynamics and Chaos by Strogatz. As part of an example, I am provided with the following electrical circuit and accompanying explanation:

Example 2.2.2: Consider the electrical circuit shown in Figure 2.2.3. A resistor \$ R \$ and a capacitor \$ C \$ are in series with a battery of constant dc voltage \$ V_0 \$. Suppose that the switch is closed at \$ t = 0 \$, and that there is no charge on the capacitor initially. Let \$ Q(t) \$ denote the charge on the capacitor at time \$ t \ge 0 \$. Sketch the graph of \$ Q(t) \$.

Solution: This type of circuit problem is probably familiar to you. It is governed by linear equations and can be solved analytically, but we prefer to illustrate the geometric approach.

First we write the circuit equations. As we go around the circuit, the total voltage drop must equal zero; hence \$ -V_0 + RI + Q/C = 0 \$, where \$ I \$ is the current flowing through the resistor. This current causes charge to accumulate on the capacitor at a rate \$ \dot{Q} = I \$. Hence

$$ -V_0 + RQ + Q/C = 0 $$

or

$$\dot{Q} = f(Q) = \dfrac{V_0}{R} - \dfrac{Q}{RC}.$$

I have started studying electronics, but I'm still learning the basics, and haven't yet learned enough to understand the electronics reasoning for what the author has written. Specifically, I'm wondering about the following:

1. Kirchhoff's voltage law states that the directed sum of the voltages around a circuit equal zero, but what is the reason for the terms of the equation \$ -V_0 + RI + Q/C = 0 \$ being structured as they are? Specifically, why is the voltage given a negative sign, why is the resistance multiplied by the current, and why is the charge on the capacitor divided by the capacitance?

2. I understand that \$ -V_0 + RI + Q/C = 0 \Rightarrow \dot{Q} = f(Q) = \dfrac{V_0}{R} - \dfrac{Q}{RC} \$ by algebra, but why is this equal to \$ f(Q) \$? In other words, why is the rate at which charge accumulates on the capacitor, \$ \dot{Q} \$, a function of the charge of the capacitor, \$ f(Q) \$?

I would greatly appreciate it if people would please take the time to clarify these points.