Physics Lesson 16.12.5 - What happens to the Current when the Source is removed from the Circuit

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Welcome to our Physics lesson on What happens to the Current when the Source is removed from the Circuit, this is the fifth lesson of our suite of physics lessons covering the topic of RL Circuits, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

What Happens to the Current when the Source is Removed from the Circuit?

When we suddenly remove the battery from a RL circuit, the current does not drop immediately to zero, as the charges in the circuit drop at a rate given by the equation

i(t) = i₀ ∙ e^{- t/τ_L}

where τ_L = L/R as explained earlier. The equation above means the current drops according an asymptotic fashion, in the same way as it rises when turning on the switch in presence of a steady source, discussed earlier in the solved example.

Let's consider another example to explain this point.

Example 2

The switch of the RL circuit shown in the figure has been turned ON for a long time. Suddenly, someone turns the switch OFF.

Calculate:

Physics Tutorials: This image provides visual information for the physics tutorial RL Circuits

The initial value of current in the circuit
The value of current in the circuit 0.02s after the switch turns OFF
The potential differences across the resistor and inductor 0.02 s after the switch turns OFF
The value of current in the circuit 20 s after the switch turns OFF.

Solution 2

We must use the equation
i(t) = ε/R ∙ e^-t/τ_L
to calculate the value of current in a RL circuit during a current drop due to the power source removal. We have the following clues:
ε = 48 V
R = 12 Ω
L = 5 H
Initially we have t = 0. Substituting these values in the equation
i(0) = ε/R ∙ e^{- 0/τ_L}
where τ_L = L/R, we obtain
i(0) = (/48/12 A) ∙ e^{- 12 ∙ 0/5}
= (4A) ∙ e⁰
= (4A) ∙ 1
= 4 A
We can find this value in an easier way by applying the Ohm's Law, because when the circuit is operating since a long time, the inductor behaves as a conducting wire. In this case, no exponential term would exist in the equation of current in the circuit, i.e.
i(0) = ε/R
= 48 V/12 Ω
= 4 A
For t = 0.02 s after the switch turns OFF, we obtain for the current in the circuit
i(0) = (48/12 A) ∙ e^{- 12 ∙ 0.02/5}
= (4A) ∙ e^{- 0.048}
= (4A) ∙ 0.953
= 3.812 A
This value is slightly lower than the value of current before the switch is turned OFF.
The potential difference across the resistor at t = 0.02 s is
ΔV_r (t) = i(t) ∙ R
ΔV_r (0.02) = i(0.02) ∙ R
= (3.812 A) ∙ (12 Ω)
= 45.744 V
The potential difference across the inductor at t = 0.02 s, is
ΔV_L (t) = ε - ΔV_r (t)
= 48 V - 45.744 V
= 2.256 V
Since τ_L = L/R = 5/12, we obtain for t = 20s:
i(t) = ε/R ∙ e^{- t/τ_L}
i(20) = 48/12 ∙ e^{- 12 ∙ 20/5}
= (4 A) ∙ e^{- 48}
= (4A) ∙ (1.425 × 10^-21)
This value is very close to zero because 1.425 × 10^-21 is a very small number. Hence, the current in the circuit 20 seconds after the switch is turned OFF, is practically zero.

How to Find the Time in which the current in a RC Circuit Reaches a Given Faction of Initial or Maximum Current?

Sometimes, we are required to find the time when a certain part of the initial or maximum value of current is flowing through the circuit. Let's explain this point through an example.

Example 3

A conductor has an inductance of 500 mH and it is connected in series to a 5Ω resistor. If the switch turns ON, how long will take to the current in the circuit to reach 80% of its maximum value?

Solution 3

The initial current in the circuit is zero and then, it starts increasing until it reaches the maximum value obtained through the Ohm's law. Therefore, we must use the equation

i(t) = ε/R ∙ (1 - e^{- R ∙ t/L} )

to calculate the current flowing in the circuit at any instant t. Giving that

L = 500 mH = 0.5 H
R = 5 Ω
i(t) = 80% of i^max = 0.8 ∙ i^max

and since

i_max = ε/R

we calculate the time required for this process after making the substitutions:

0.8 ∙ i_max = i_max ∙ 1 - e^{- 5 ∙ t/0.5}

Simplifying i_max from both sides, we obtain

0.8 = 1 - e^-10t
e^-10t = 1-0.8
e^-10t = 0.2
ln⁡(e^-10t ) = ln⁡0.2
-10t = -1.61
t = -1.61/-10
= 0.161 s

Thus, the current in the circuit will reach 80% of its maximum value 0.161 s after turning the switch ON.

You have reached the end of Physics lesson 16.12.5 What happens to the Current when the Source is removed from the Circuit. There are 5 lessons in this physics tutorial covering RL Circuits, you can access all the lessons from this tutorial below.

More RL Circuits Lessons and Learning Resources

Magnetism Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
16.12	RL Circuits
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
16.12.1	What Are RL Circuits? An Overview
16.12.2	A Short Recap on RC Circuits as a Useful Step to Understand the RL Circuits
16.12.3	RL Circuits
16.12.4	What is the Dimension of τ?
16.12.5	What happens to the Current when the Source is removed from the Circuit

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