In the circuit shown in Fig. 30.27, ℰ=60.0 V, R1= 40.0 Ω, R2=25.0 Ω, and L= 0.300 H . Switch S

In the circuit shown in Fig. 30.27, ℰ=60.0 V, R1= 40.0 Ω,...

In the circuit shown in Fig. $30.27, \quad \mathcal{E}=60.0 \mathrm{V}, R_{1}=$ $40.0 \Omega, R_{2}=25.0 \Omega,$ and $L=$ $0.300 \mathrm{H} .$ Switch $\mathrm{S}$ is closed at $t=0 .$ Just after the switch is closed, (a) what is the potential difference $v_{a b}$ across the resistor $R_{1} ;(b)$ which point, $a$ or $b,$ is at a higher potential; (c) what is the potential difference $v_{c d}$ across the inductor $L ;$ (d) which point, $c$ or $d,$ is at a higher potential? The switch is left closed a long time and then opened. Just after the switch is opened, (e) what is the potential difference $v_{a b}$ across the resistor $R_{1} ;(f)$ which point, $a$ or $b,$ is at a higher potential; (g) what is the potential difference $v_{c d}$ across the inductor $L_{i}$ (h) which point, $c$ or $d,$ is at a higher potential?

Key Concepts

Transient Response in RL Circuits

When a circuit containing a resistor (R) and an inductor (L) is switched on or off, the current does not instantly settle into its new value but changes over time. This time-dependent change, known as a transient response, is governed by differential equations that account for the time constant of the circuit. The rate at which the current approaches its steady state is determined by the product of the inductance and resistance (? = L/R), which dictates how quickly the inductor’s magnetic field builds or collapses in response to changes in current.

Inductor Behavior and Back EMF

Inductors have the key property of opposing changes in current flowing through them. This opposition is manifested as an induced electromotive force (EMF), often referred to as back EMF, which is proportional to the rate of change of current (V = L(di/dt)). In transient states, such as immediately after a switch is closed or opened, the inductor’s back EMF plays a crucial role in determining how voltages distribute across circuit elements. The inductor resists sudden changes by momentarily acting like an open circuit when the current is just beginning to change.

Steady-State Behavior in DC Circuits

In DC circuits, once enough time has elapsed after a switching event, the transient effects subside and the circuit reaches a steady state. At steady state, the inductor behaves effectively as a short circuit because the current becomes constant, resulting in no change in current (di/dt = 0) and, consequently, no voltage drop across the inductor. Understanding this behavior is fundamental when analyzing circuits over long times after a switching event.

Kirchhoff's Voltage Law (KVL)

Kirchhoff's Voltage Law is a fundamental principle for analyzing electrical circuits. It states that the sum of the electric potential differences around any closed loop in a circuit must equal zero. This law is consistently applied in transient and steady-state analysis to ensure that all voltage gains (such as from batteries) and voltage drops (across resistors, inductors, etc.) are correctly accounted for when determining the potential differences at various points in the circuit.

Transcript

00:01 In the first part of this problem, we are going to calculate the potential difference, v -a -b.

00:07 So let's write the equation for this potential difference.

00:11 We can write e minus v -a -b is equal to 0 -volt.

00:16 So from here we can write this v -a -b is equal to e.

00:21 Now by inserting value for this e, we can write v -a -b is equal to 60 .0 -volt.

00:33 Part b of this problem, we are going to identify the point of higher potential.

00:39 The voltage drop across the resistor shows that the point is at higher potential.

00:45 In part c of this problem, we are going to calculate the potential difference across the inductor.

00:51 That is what we call v -c -d.

00:54 In order to calculate this value, let's apply the loop rule.

00:59 So we can wind e minus v -r -2 minus v -l.

01:07 We can say that vcd which is equal to 0.

01:11 Now here we know that this vr2 is equals to i2 and this is equal to 0 in this case so we can write this e minus 0 minus vcd is equal to 0.

01:31 So from here we can write vcd is equal to 0.

01:31 So from here we can write vcd is equal to e so by inserting value of e, into this equation we can write vcd is equal to 60 .0 volt.

01:47 In party of this problem we have to identify the point of higher potential.

01:52 As we move from b to a the voltage increases so the voltage must drop if we move in opposite direction which is from a to b.

02:01 This shows that point c must be at higher potential.

02:05 In part e of this problem we are going to calculate the potential difference across the register 1 that is vab.

02:17 This is vab.

02:20 So let's apply the loop rule...

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