Consider the circuit shown in Fig. 30.28 . Let ℰ=36.0 V, R0= 50.0 Ω, R=150 Ω, and L=4.00 H . (a)

Consider the circuit shown in Fig. 30.28 . Let ℰ=36.0 V, R0=...

Consider the circuit shown in Fig. 30.28 . Let $\mathcal{E}=36.0 \mathrm{V}, R_{0}=$ $50.0 \Omega, R=150 \Omega,$ and $L=4.00 \mathrm{H} .$ (a) Switch $S_{1}$ is closed and switch $S_{2}$ is left open. Just after $S_{1}$ is closed, what are the current $i_{0}$ through $R_{0}$ and the potential differences $v_{a c}$ and $v_{c b}$ ? (b) After $S_{1}$ has been closed a long time $\left(\mathrm{S}_{2} \text { is still open) so that the cur- }\right.$ rent has reached its final, steady value, what are $i_{0}, v_{a c},$ and $v_{c b} ?$ (c) Find the expressions for $i_{0}, v_{\alpha c},$ and $v_{c b}$ as functions of the time $t$ since $S_{1}$ was closed. Your results should agree with part (a) when $t=0$ and with part $(b)$ when $t \rightarrow \infty$ . Graph $i_{0}, v_{a c},$ and $v_{c b}$ versus time.

Consider the circuit shown in Fig. 30.28 . Let $\mathcal{E}=36.0 \mathrm{V}, R_{0}=$ $50.0 \Omega, R=150 \Omega,$ and $L=4.00 \mathrm{H} .$ (a) Switch $S_{1}$ is closed and switch $S_{2}$ is
left open. Just after $S_{1}$ is closed, what are the current $i_{0}$ through $R_{0}$ and the potential differences $v_{a c}$ and $v_{c b}$ ? (b) After $S_{1}$ has been closed a long time $\left(\mathrm{S}_{2} \text { is still open) so that the cur- }\right.$ rent has reached its final, steady value, what are $i_{0}, v_{a c},$ and $v_{c b} ?$ (c) Find the expressions for $i_{0}, v_{\alpha c},$ and $v_{c b}$ as functions of the time $t$ since $S_{1}$ was closed. Your results should agree with part (a) when $t=0$ and with part $(b)$ when $t \rightarrow \infty$ . Graph $i_{0}, v_{a c},$ and $v_{c b}$ versus time.

Key Concepts

RL Circuit Transients

This concept involves understanding how circuits containing inductors and resistors respond immediately after a switching event. When a switch is closed in an RL circuit, the current does not instantly reach its steady-state value because the inductor resists changes in current. Therefore, the transient response is characterized by a time-dependent change in current and voltages, typically described by an exponential function governed by the circuit's time constant.

Inductor Behavior

Inductors play a key role in transient circuits because they resist abrupt changes in current. When the circuit is first energized, an ideal inductor behaves like an open circuit, preventing current from immediately flowing. Over time, the inductor gradually allows current to increase until it reaches steady-state conditions. This behavior is essential for understanding the initial and evolving voltage differences across circuit elements in the transient analysis.

Steady-State Analysis in RL Circuits

Steady-state analysis deals with the behavior of a circuit after a long period has elapsed since any switching action. In an RL circuit, once the transient effects have decayed, the inductor behaves like a short circuit, allowing the current to be determined solely by the resistive components. Analyzing the circuit under steady-state conditions simplifies the calculations, as the time-varying terms drop out and the circuit can be treated using conventional Ohm’s law and Kirchhoff’s laws.

Differential Equation Modeling

The evolution of current and voltage in an RL circuit is mathematically described by differential equations that incorporate the inductor’s property of opposing changes in current. These equations typically result in solutions with exponential behavior, reflecting the gradual transition from initial conditions to steady state. Mastering the setup and solution of these differential equations is crucial for accurately predicting the transient response of the circuit.

Time Constant in RL Circuits

The time constant, generally defined as ? = L/R for an RL circuit, quantifies the timescale over which the transient effects decay. It indicates how rapidly the circuit transitions from its initial response immediately after a switch action to its final steady-state behavior. A larger time constant implies a slower response, while a smaller time constant indicates a faster approach to steady state.

Transcript

00:01 In part of this problem we are going to calculate i not then we have to write the value for v a c and vc b.

00:15 Now at t equals to 0 seconds we know that this i not will be equals to 0 ampere and we can write vac c as v a c is equals to i0 r not so we can write 0 ampere into r0.

00:38 So this will give the value for this vac as 0.

00:43 Now using the loop rule, we can write e minus vace minus vacb equals to 0.

00:56 So from here we can write the value for this vcb as vcb is equal to e minus vacc.

01:06 Now by inserting values into this square we can write v -cb is equal to 36 .0 -volt minus 0 -volt.

01:16 So this will give the value for this v -c -b as v -cb is equal to 36 .0.

01:26 In part b of this problem, we are going to calculate the value of i -0, then value of vacc and vcb at t goes to infinity.

01:47 So when t goes to infinity, at that time this d i .0 divided by dt will be equals to 0.

01:57 So this lead to the problem that el is equals to minus l, d i .0 divided by dt and this will be equal to 0.

02:10 Now, by using loop rule, we can write this e minus i0 into r0 plus r is equals to 0.

02:23 From here we can write the value for this i note as i0 is equal to e divided by r0 plus r so by inserting values into this square you can write 36 .0, divided by 50.

02:40 0 .0 omega plus 150 omega.

02:47 So this will give the value for this i not as i not is equals to 0 .180 amp.

02:57 Now we can write vse as vse is equals to i0rr.

03:05 So this will be equals to 0 .180 amp into r not which is equal to 50.

03:13 0 omega.

03:15 So this will give you a value for this vse as 9 .0 volt.

03:21 Now we can write the equation for vcb as vcb is equal to i .0 r plus el.

03:38 So from here we can write as vcb is equal to 0 .180 ampere into 1 .80 150 omega plus 0.

03:54 So this will give you the value for this vcb as vcb is equals to 27 .0.

04:08 In part c of this problem we are going to derive the expressions for i -0, v -cb, vac, and v -cb.

04:26 And we have to plot the graphs...