Consider a series circuit (Figure 4) consisting of a...

Consider a series circuit (Figure 4$)$ consisting of a resistor of $R$ ohms, an inductor of $L$ henries, and a variable voltage source of $V(t)$ volts (time t in seconds). The current through the circuit $I(t)$ (in amperes) satisfes the differential equation $$ \frac{d I}{d t}+\frac{R}{L} I=\frac{1}{L} V(t) $$ Find the solution to Eq. ( 10 ) with initial condition $I(0)=0,$ assuming that $R=100 \Omega, L=5 \mathrm{H},$ and $V(t)$ is constant with $V(t)=10 \mathrm{V}$ .

Consider a series circuit (Figure 4$)$ consisting of a resistor of $R$ ohms, an inductor of $L$ henries, and a variable voltage source of $V(t)$ volts (time t in seconds). The current through the circuit $I(t)$ (in amperes) satisfes the differential equation
$$
\frac{d I}{d t}+\frac{R}{L} I=\frac{1}{L} V(t)
$$
Find the solution to Eq. ( 10 ) with initial condition $I(0)=0,$ assuming that $R=100 \Omega, L=5 \mathrm{H},$ and $V(t)$ is constant with $V(t)=10 \mathrm{V}$ .

Key Concepts

Transient and Steady-State Response

In the context of electrical circuits, the transient response refers to the temporary behavior of the circuit immediately after a change in conditions, typically characterized by exponential decay or growth. The steady-state response is the long-term behavior once all transients have died out, often corresponding to a constant value determined by the circuit's parameters.

RL Circuit Analysis

Analyzing an RL (resistor-inductor) circuit involves setting up and solving differential equations that describe how current changes over time due to the resistive and inductive properties. The key parameters include the resistance, inductance, and any applied voltage, which determine both the transient and steady-state behavior of the circuit.

First Order Linear Differential Equations

This concept involves equations that describe the rate of change of a variable and can be written in the form dy/dt + P(t)y = Q(t). The solution process typically includes solving for the homogeneous solution and finding a particular solution, which together satisfy any specified initial conditions.

Integrating Factor Method

This is a standard technique used to solve first order linear differential equations. It involves multiplying the entire equation by an integrating factor, usually of the form e^(?P(t) dt), which facilitates rewriting the left-hand side of the equation as the derivative of a product, making the equation easier to integrate.

Transcript

00:01 All right, we are looking at a series circuit.

00:03 Now, if you don't know what a series circuit is, pretty much you've got a single pathway that carries a current.

00:13 Where your r is your resistor, your l in this case is an inductor, and your vt is your variable volts, and your time here is measured in seconds.

00:24 Now, we do have one equation that we're using here.

00:27 It does have to satisfy our current equation.

00:31 D -i over d -t plus r over l -5 is equal to 1 over l -d -t.

00:42 So we are satisfying that equation.

00:45 Now for this case, they've told us that our i at 0 is 0.

00:51 Our resistance is a hundred.

00:57 Our inductor is 5 emiris.

01:04 And our variable voltage is a constant here.

01:07 At 10 volts and that's going to be constant.

01:12 So they want us to try to solve this equation.

01:16 So all we're going to do is we're going to substitute in what we know.

01:21 So we know that r is 100, l is 5, 1 over 5 times, and cleaning that up with it, we end up with 100 divided by 5, that is 20, and 1 5 times 10 is 2.

01:43 So what we see now is we kind of have this equation here.

01:47 If i wanted to, i could actually change out this bia over dt, and instead i'm going to write it as i prime.

01:54 Because now it looks like our general equation that we would use to do a first order derivative.

02:06 So once we have that, we would know that our a function is 20.

02:12 Our b function is 2.

02:16 And to find out our lowercase a function, we would go e the power of the integral of 20, which happens to be e the power of 1t.

02:28 Once we've got that, we can substitute it into our equation...