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23.10 Rl circuits  (Page 2/3)

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I = I 0 e − t / τ     (turning off). size 12{I=I rSub { size 8{0} } e rSup { size 8{ - t/τ} } } {}

(See [link] (c).) In the first period of time τ = L / R size 12{τ=L/R} {} after the switch is closed, the current falls to 0.368 of its initial value, since I = I 0 e − 1 = 0 . 368 I 0 size 12{I=I rSub { size 8{0} } e rSup { size 8{ - 1} } =0 "." "368"I rSub { size 8{0} } } {} . In each successive time τ size 12{τ} {} , the current falls to 0.368 of the preceding value, and in a few multiples of τ size 12{τ} {} , the current becomes very close to zero, as seen in the graph in [link] (c).

Calculating characteristic time and current in an RL Circuit

(a) What is the characteristic time constant for a 7.50 mH inductor in series with a 3.00 Ω resistor? (b) Find the current 5.00 ms after the switch is moved to position 2 to disconnect the battery, if it is initially 10.0 A.

Strategy for (a)

The time constant for an RL circuit is defined by τ = L / R size 12{τ=L/R} {} .

Solution for (a)

Entering known values into the expression for τ size 12{τ} {} given in τ = L / R size 12{τ=L/R} {} yields

τ = L R = 7.50 mH 3.00 Ω = 2.50 ms. size 12{τ= { {L} over {R} } = { {7 "." "50"" mH"} over {3 "." "00 " %OMEGA } } =2 "." "50"" ms"} {}

Discussion for (a)

This is a small but definitely finite time. The coil will be very close to its full current in about ten time constants, or about 25 ms.

Strategy for (b)

We can find the current by using I = I 0 e − t / τ size 12{I=I rSub { size 8{0} } e rSup { size 8{ - t/τ} } } {} , or by considering the decline in steps. Since the time is twice the characteristic time, we consider the process in steps.

Solution for (b)

In the first 2.50 ms, the current declines to 0.368 of its initial value, which is

I = 0 . 368 I 0 = ( 0 . 368 ) ( 10.0 A ) = 3 . 68 A at  t = 2 . 50  ms.

After another 2.50 ms, or a total of 5.00 ms, the current declines to 0.368 of the value just found. That is,

I ′ = 0 . 368 I = ( 0 . 368 ) ( 3.68 A ) = 1 . 35  A at  t = 5 . 00  ms. alignl { stack { size 12{ { {I}} sup { ' }=0 "." "368"I= \( 0 "." "368" \) \( 3 "." "68"" A" \) } {} #size 12{" "=1 "." "35"" A at "t=5 "." "00"" ms"} {} } } {}

Discussion for (b)

After another 5.00 ms has passed, the current will be 0.183 A (see [link] ); so, although it does die out, the current certainly does not go to zero instantaneously.

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In summary, when the voltage applied to an inductor is changed, the current also changes, but the change in current lags the change in voltage in an RL circuit . In Reactance, Inductive and Capacitive , we explore how an RL circuit behaves when a sinusoidal AC voltage is applied.

Section 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 = I 0 ( 1 − e − t / τ )     (turning on). size 12{I=I rSub { size 8{0} } \( 1 - e rSup { size 8{ - t/τ} } \) } {}
    where I 0 = V / R size 12{I rSub { size 8{0} } =V/R} {} is the final current.
  • The characteristic time constant τ size 12{τ} {} is τ = L R size 12{τ= { {L} over {R} } } {} , where L is the inductance and R is the resistance.
  • In the first time constant τ size 12{τ} {} , the current rises from zero to 0 . 632 I 0 size 12{0 "." "632"I rSub { size 8{0} } } {} , and 0.632 of the remainder in every subsequent time interval τ size 12{τ} {} .
  • When the inductor is shorted through a resistor, current decreases as
    I = I 0 e − t / τ     (turning off). size 12{I=I rSub { size 8{0} } e rSup { size 8{ - t/τ} } } {}
    Here I 0 size 12{I rSub { size 8{0} } } {} is the initial current.
  • Current falls to 0 . 368 I 0 size 12{0 "." "368"I rSub { size 8{0} } } {} in the first time interval τ size 12{τ} {} , and 0.368 of the remainder toward zero in each subsequent time τ size 12{τ} {} .

Problem exercises

If you want a characteristic RL time constant of 1.00 s, and you have a 500 Ω resistor, what value of self-inductance is needed?

500 H

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Your RL circuit has a characteristic time constant of 20.0 ns, and a resistance of 5.00 MΩ . (a) What is the inductance of the circuit? (b) What resistance would give you a 1.00 ns time constant, perhaps needed for quick response in an oscilloscope?

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A large superconducting magnet, used for magnetic resonance imaging, has a 50.0 H inductance. If you want current through it to be adjustable with a 1.00 s characteristic time constant, what is the minimum resistance of system?

50.0 Ω

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Verify that after a time of 10.0 ms, the current for the situation considered in [link] will be 0.183 A as stated.

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Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and resistors ranging from 0.100 Ω to 1.00 MΩ . What is the range of characteristic RL time constants you can produce by connecting a single resistor to a single inductor?

1 . 00 × 10 –18 s size 12{1 "." "00" times "10" rSup { size 8{"-15"} } " s"} {} to 0.100 s

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(a) What is the characteristic time constant of a 25.0 mH inductor that has a resistance of 4.00 Ω ? (b) If it is connected to a 12.0 V battery, what is the current after 12.5 ms?

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What percentage of the final current I 0 flows through an inductor L size 12{L} {} in series with a resistor R size 12{R} {} , three time constants after the circuit is completed?

95.0%

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The 5.00 A current through a 1.50 H inductor is dissipated by a 2.00 Ω resistor in a circuit like that in [link] with the switch in position 2. (a) What is the initial energy in the inductor? (b) How long will it take the current to decline to 5.00% of its initial value? (c) Calculate the average power dissipated, and compare it with the initial power dissipated by the resistor.

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(a) Use the exact exponential treatment to find how much time is required to bring the current through an 80.0 mH inductor in series with a 15.0 Ω resistor to 99.0% of its final value, starting from zero. (b) Compare your answer to the approximate treatment using integral numbers of τ size 12{τ} {} . (c) Discuss how significant the difference is.

(a) 24.6 ms

(b) 26.7 ms

(c) 9% difference, which is greater than the inherent uncertainty in the given parameters.

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(a) Using the exact exponential treatment, find the time required for the current through a 2.00 H inductor in series with a 0.500 Ω resistor to be reduced to 0.100% of its original value. (b) Compare your answer to the approximate treatment using integral numbers of τ size 12{τ} {} . (c) Discuss how significant the difference is.

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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