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Check Your Understanding Verify that RC and L/R have the dimensions of time.

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Check Your Understanding (a) If the current in the circuit of in [link] (b) increases to 90 % of its final value after 5.0 s, what is the inductive time constant? (b) If R = 20 Ω , what is the value of the self-inductance? (c) If the 20 - Ω resistor is replaced with a 100 - Ω resister, what is the time taken for the current to reach 90 % of its final value?

a. 2.2 s; b. 43 H; c. 1.0 s

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Check Your Understanding For the circuit of in [link] (b), show that when steady state is reached, the difference in the total energies produced by the battery and dissipated in the resistor is equal to the energy stored in the magnetic field of the coil.

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Summary

  • When a series connection of a resistor and an inductor—an RL circuit—is connected to a voltage source, the time variation of the current is
    I ( t ) = ε R ( 1 e R t / L ) = ε R ( 1 e t / τ L ) (turning on),
    where the initial current is I 0 = ε / R .
  • The characteristic time constant τ is τ L = L / R , where L is the inductance and R is the resistance.
  • In the first time constant τ , the current rises from zero to 0.632 I 0 , and to 0.632 of the remainder in every subsequent time interval τ .
  • When the inductor is shorted through a resistor, current decreases as
    I ( t ) = ε R e t / τ L (turning off).
    Current falls to 0.368 I 0 in the first time interval τ , and to 0.368 of the remainder toward zero in each subsequent time τ .

Conceptual questions

Use Lenz’s law to explain why the initial current in the RL circuit of [link] (b) is zero.

As current flows through the inductor, there is a back current by Lenz’s law that is created to keep the net current at zero amps, the initial current.

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When the current in the RL circuit of [link] (b) reaches its final value ε / R , what is the voltage across the inductor? Across the resistor?

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Does the time required for the current in an RL circuit to reach any fraction of its steady-state value depend on the emf of the battery?

no

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An inductor is connected across the terminals of a battery. Does the current that eventually flows through the inductor depend on the internal resistance of the battery? Does the time required for the current to reach its final value depend on this resistance?

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At what time is the voltage across the inductor of the RL circuit of [link] (b) a maximum?

At t = 0 , or when the switch is first thrown.

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In the simple RL circuit of [link] (b), can the emf induced across the inductor ever be greater than the emf of the battery used to produce the current?

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If the emf of the battery of [link] (b) is reduced by a factor of 2, by how much does the steady-state energy stored in the magnetic field of the inductor change?

1/4

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A steady current flows through a circuit with a large inductive time constant. When a switch in the circuit is opened, a large spark occurs across the terminals of the switch. Explain.

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Describe how the currents through R 1 and R 2 shown below vary with time after switch S is closed.

Figure shows a circuit with resistor R1 connected in series with battery epsilon, through open switch S. R1 is parallel to resistor R2 and inductor L.

Initially, I R 1 = ε R 1 and I R 2 = 0 , and after a long time has passed, I R 1 = ε R 1 and I R 2 = ε R 2 .

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Discuss possible practical applications of RL circuits.

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Problems

In [link] , ε = 12 V , L = 20 mH , and R = 5.0 Ω . Determine (a) the time constant of the circuit, (b) the initial current through the resistor, (c) the final current through the resistor, (d) the current through the resistor when t = 2 τ L , and (e) the voltages across the inductor and the resistor when t = 2 τ L .

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For the circuit shown below, ε = 20 V , L = 4.0 mH, and R = 5.0 Ω . After steady state is reached with S 1 closed and S 2 open, S 2 is closed and immediately thereafter ( at t = 0 ) S 1 is opened. Determine (a) the current through L at t = 0 , (b) the current through L at t = 4.0 × 10 −4 s , and (c) the voltages across L and R at t = 4.0 × 10 −4 s .

Figure shows a circuit with R and L connected in series with battery epsilon through closed switch S. L is connected in parallel with another resistor R through open switch S2.

a. 4.0 A; b. 2.4 A; c. on R : V = 12 V ; on L : V = 7.9 V

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The current in the RL circuit shown here increases to 40 % of its steady-state value in 2.0 s. What is the time constant of the circuit?

Figure a shows a resistor R and an inductor L connected in series with two switches which are parallel to each other. Both switches are currently open. Closing switch S1 would connect R and L in series with a battery, whose positive terminal is towards L. Closing switch S2 would form a closed loop of R and L, without the battery. Figure b shows a closed circuit with R, L and the battery in series. The side of L towards the battery, is at positive potential. Current flows from the positive end of L, through it, to the negative end. Figure c shows R and L connected in series. The potential across L is reversed, but the current flows in the same direction as in figure b.
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How long after switch S 1 is thrown does it take the current in the circuit shown to reach half its maximum value? Express your answer in terms of the time constant of the circuit.

Figure shows a circuit with R and L in series with a battery, epsilon and a switch S1 which is open.

0.69 τ

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Examine the circuit shown below in part (a). Determine dI/dt at the instant after the switch is thrown in the circuit of (a), thereby producing the circuit of (b). Show that if I were to continue to increase at this initial rate, it would reach its maximum ε / R in one time constant.

Figure a shows a circuit with R and L in series with a battery, epsilon and a switch S1 which is open. Figure b shows a circuit with R and L in series with a battery, epsilon. The end of L that is connected to the positive terminal of the battery is at positive potential. Current flows through L from the positive end to the negative one.
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The current in the RL circuit shown below reaches half its maximum value in 1.75 ms after the switch S 1 is thrown. Determine (a) the time constant of the circuit and (b) the resistance of the circuit if L = 250 mH .

Figure shows a circuit with R and L in series with a battery, epsilon and a switch S1 which is open.

a. 2.52 ms; b. 99.2 Ω

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Consider the circuit shown below. Find I 1 , I 2 , and I 3 when (a) the switch S is first closed, (b) after the currents have reached steady-state values, and (c) at the instant the switch is reopened (after being closed for a long time).

Figure shows a circuit with R1 and L connected in series with a battery epsilon and a closed switch S. R2 is connected in parallel with L. The currents through R1, L and R2 are I1, I2 and I3 respectively.
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For the circuit shown below, ε = 50 V , R 1 = 10 Ω, and L = 2.0 mH . Find the values of I 1 and I 2 (a) immediately after switch S is closed, (b) a long time after S is closed, (c) immediately after S is reopened, and (d) a long time after S is reopened.

Figure shows a circuit with R1 and R2 connected in series with a battery, epsilon and a closed switch S. R2 is connected in parallel with L and R3. The currents through R1 and R2 are I1 and I2 respectively.

a. I 1 = I 2 = 1.7 A ; b. I 1 = 2.73 A , I 2 = 1.36 A ; c. I 1 = 0 , I 2 = 0.54 A ; d. I 1 = I 2 = 0

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For the circuit shown below, find the current through the inductor 2.0 × 10 −5 s after the switch is reopened.

Figure shows a circuit with R1 and R2 connected in series with a battery, epsilon and a closed switch S. R2 is connected in parallel with L and R3. The currents through R1 and R2 are I1 and I2 respectively.
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Show that for the circuit shown below, the initial energy stored in the inductor, L I 2 ( 0 ) / 2 , is equal to the total energy eventually dissipated in the resistor, 0 I 2 ( t ) R d t .

Figure a shows a resistor R and an inductor L connected in series with two switches which are parallel to each other. Both switches are currently open. Closing switch S1 would connect R and L in series with a battery, whose positive terminal is towards L. Closing switch S2 would form a closed loop of R and L, without the battery. Figure b shows a closed circuit with R, L and the battery in series. The side of L towards the battery, is at positive potential. Current flows from the positive end of L, through it, to the negative end. Figure c shows R and L connected in series. The potential across L is reversed, but the current flows in the same direction as in figure b.

proof

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Questions & Answers

a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
Abu Reply
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
Aditya
determine absolute zero
OFERE Reply
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
Ashok Reply
what is the weight of the earth in space
peterpaul Reply
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
Ifeoma Reply
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
Mehmet
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
Mateshwar Reply
explain the lack of symmetry in the field of the parallel capacitor
Phoebe Reply
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
does your app come with video lessons?
Ahmed Reply
What is vector
Ajibola Reply
Vector is a quantity having a direction as well as magnitude
Damilare
Practice Key Terms 1

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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