A 9.00-V battery is connected through a switch to two...

A $9.00-V$ battery is connected through a switch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors has a resistance of $100 . \Omega$, and the inductor has an inductance of $3.00 \mathrm{H}$. The switch is initially open. a) Immediately after the switch is closed, what is the current in resistor $R_{1}$ and in resistor $R_{2}$ ? b) At $50.0 \mathrm{~ms}$ after the switch is closed, what is the current in resistor $R_{1}$ and in resistor $R_{2} ?$ c) At $500 . \mathrm{ms}$ after the switch is closed, what is the current in resistor $R_{1}$ and in resistor $R_{2} ?$ d) After a long time $(>10.0 \mathrm{~s})$, the switch is opened again. Immediately after the switch is opened, what is the current in resistor $R_{1}$ and in resistor $R_{2}$ ? e) At $50.0 \mathrm{~ms}$ after the switch is opened, what is the current in resistor $R_{1}$ and in resistor $R_{2} ?$ f) At $500 . \mathrm{ms}$ after the switch is opened, what is the current in resistor $R_{1}$ and in resistor $R_{2} ?$

A $9.00-V$ battery is connected through a switch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors has a resistance of $100 . \Omega$, and the inductor has an inductance of $3.00 \mathrm{H}$. The switch is initially open.
a) Immediately after the switch is closed, what is the current in resistor $R_{1}$ and in resistor $R_{2}$ ?
b) At $50.0 \mathrm{~ms}$ after the switch is closed, what is the current in resistor $R_{1}$ and in resistor $R_{2} ?$
c) At $500 . \mathrm{ms}$ after the switch is closed, what is the current in resistor $R_{1}$ and in resistor $R_{2} ?$
d) After a long time $(>10.0 \mathrm{~s})$, the switch is opened again. Immediately after the switch is opened, what is the current in resistor $R_{1}$ and in resistor $R_{2}$ ?
e) At $50.0 \mathrm{~ms}$ after the switch is opened, what is the current in resistor $R_{1}$ and in resistor $R_{2} ?$
f) At $500 . \mathrm{ms}$ after the switch is opened, what is the current in resistor $R_{1}$ and in resistor $R_{2} ?$

Key Concepts

Steady-State Conditions in RL Circuits

After a long period following a switching event, the circuit reaches steady state where the current becomes constant. In this state, the voltage across the inductor drops to zero, and it behaves as a short circuit for DC analysis. Understanding the steady state is critical for determining the long-term behavior of the circuit and the eventual current distribution among circuit elements.

Kirchhoff's Circuit Laws

Kirchhoff's voltage and current laws are fundamental to analyzing circuits containing resistors and inductors. These laws allow the establishment of relationships between the voltages and currents in different branches of the circuit, providing a systematic way to derive both transient and steady-state equations necessary for solving complex circuit problems.

Inductor Behavior at Switching

Inductors have the intrinsic property of opposing changes in current due to their stored magnetic energy. This results in the inductor acting like an open circuit immediately after a switch is closed—preventing an abrupt change in current—and behaving like a short circuit after a long time when steady state is reached. The same principle applies when the circuit is opened, as the inductor attempts to maintain its current initially.

RL Circuit Transients

This concept refers to the changing behavior of current and voltage in a circuit immediately after a switching event, before the system settles into a steady state. Transient analysis in RL circuits involves understanding how current starts, builds up, or decays exponentially over time, and is crucial when predicting circuit behavior during switching operations.

Time Constant

The time constant in an RL circuit, defined as ? = L/R, is a measure of the rate at which the circuit reaches its steady state. It quantifies how quickly the current through the inductor reaches approximately 63% of its final value during growth or decays to about 37% of its initial value during discharge, providing a key indicator for the response speed of the circuit.

Exponential Growth and Decay

In RL circuits, the current does not change instantly but follows an exponential trend according to the circuit's time constant. When the circuit is energized, the current rises exponentially, and when the circuit is de-energized, the current decays exponentially. This behavior arises from the differential equations governing RL circuits and is key to predicting both transient response and steady state.

Transcript

00:01 The problem gives us a circuit with a battery, two resistors, and an inductor, as shown here.

00:10 Where v is equal to 9 volts, r1 and r2 are both 100 oms, and the inductance of the inductor is 3 henries.

00:19 Now, for question a, we are asked to find for the current, passing through r1 and r2, where the switch is immediately closed or time is equal to 0 seconds.

00:38 Now when the switch is closed, we could find for i1 using omslow.

00:48 Since the voltage source and r1 are both connected in parallel, we could say that v is equal to vr1.

01:01 Using oms law, ir1 is simply equal to vr1 over the resistance r1.

01:11 Substituting the known values will have 9 .00 volts over 100 oms.

01:19 That gives us a current of 0 .0 900 amperors.

01:30 Now to get for the current i2, we know that a resistance turn inductor in cilles can be described by the equation i2 of t is equal to the voltage over r2 times 1 minus e raised to negative r t over l now since the case dictates that time equals 0 and substituting the known values we could find for the current i2 as 9 .0 volts over the resistance r2, which is 100 oms, times 1 minus e raised to negative r times 0 over l.

02:26 So the exponent here is simply 0 and any number raised to 0 is just equal to 1.

02:35 Hence, we'll arrive at 9 .0 volts over 100 oms times 1 minus 1, or 0.

02:47 Therefore, current passing through r2 is equal to 0 just after the switch is closed.

03:03 Now for question letter b, we are asked to find for the current in r1 and r2, at time equals 50 .0 milliseconds after the switch was closed.

03:20 So we'll use the same current equation that is i2 of t is equal to 9 .0 volts over 100 oms times 1 minus negative 100 oms times time where 50 milliseconds is equal to 0 .00 seconds, divided by the inductance 3 henries.

04:00 Therefore, current i2 at 50 .0 milliseconds is equal to 0 .0730 amper.

04:17 Now, if you look at r1, at what i'm, 0 .0 .0 .3 amperors.

04:20 However time we may measure the current, the voltage across r1 does not change.

04:27 Therefore, the current will also remain the same.

04:33 That is, ir1 is equal to 0 .900 ampers.

04:46 Now for question letter c, the same methods will be used since we're only getting the current for time, this time, at 500 milliseconds or 0 .500 seconds.

05:06 Now using the same equation, we get i2 at 500 milliseconds equals 9 .0 volts over 100 oms times 1 minus e raised the negative 100 oms times 0 .500 seconds divided by three henries.

05:33 Now, calculating here will get i2 is equal to 0 .0 900 amperors.

05:58 Now, the same would go for i1 at time at 500 milliseconds.

06:09 Since the voltage does not change across r1, the current would also not change.

06:18 Therefore, i1 is equal to 0 .900 amperors.

06:25 Now, the next set of questions would have a different case.

06:31 So here, after closing this switch for a very long time, or let's say, 10 seconds, the switch is opened again.

06:39 Now for letter d, immediately after opening the switch, what would be the current in resistor 1 and resistor 2? now, we could simplify the circuit after opening the switch into simply a connection between r1, r2, and the inductor.

07:10 Since the left part of the circuit is no longer connected...

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