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In this Physics tutorial, you will learn:

- What are RL circuits?
- What do RL circuits have in common with RC circuits?
- How does the current changes in RL circuits?
- What happens to the voltages in each component of a RL circuit?
- How to calculate the current flowing at any instant through a RL circuit?
- What is the inductive time constant?
- How to find the time in which the current in a RC circuit reaches a given faction of initial or maximum current?

∆V(t) = ε ∙ (1-e^(- t/(R ∙ C)))

where ε is the electromotive force generated by a DC source (for example a battery). The charge stored in a capacitor as a function of time has a similar form, i.e. Q(t) = Q_{0} ∙ (1-e^(- t/(R ∙ C)))

where Q(t) is the charge stored in a capacitor at any time instant t and Q0 is the initial charge of the capacitor at the beginning of the charging process. The R ∙ C term is usually denoted by τ (or τC); it has the unit of time and shows how fast the RC circuit is charging or discharging. The discharge of a capacitor in a RC circuit is the inverse of charging process. In this case, we have a decreasing exponential function when considering the potential difference vs time variation. The equation of potential difference across a capacitor during the discharge process is ∆V(t) = ε ∙ e^(- t/(R ∙ C))

and the charge remained in the capacitor at any instant during the discharge process, is Q(t) = Q_{0} ∙ e^(- t/(R ∙ C))

Since the capacitance of a capacitor is C = Q/∆V

and if we replace the potential difference ΔV with the electromotive force ε produced by the source (if we neglect the resistance of wire, then ΔV ≈ ε), we obtain for the charge stored in capacitor in terms of capacitance C: Q(t) = C ∙ ε ∙ (1-e^(- t/(R ∙ C)))

RL Circuits The rise or fall in the number of charges in a capacitor charging through a resistor results in a rise or fall in the current in the circuit as I(t) = ∆Q/∆t

For very small time intervals dt, we can write I(t) = dQ/dt

Similarly, a rise or a fall of the current occurs when a source by an electromotive force ε supplies a single loop circuit containing a resistor R and an inductor L. When the switch S is moved at position a, the current in the resistor starts to rise. In absence of inductor, this rise of current from zero to a steady value would be immediate and we used the Ohm's law to find this rise in current, i.e. I = ε/R

When an inductor is present in the circuit however, a self-induced emf εL appears in the circuit. The current generated because of this self-induced emf is in the opposite direction of the current produced by the battery. As a result, it opposes the rise in current (from the Lentz law) and this causes a delay in the rise of current in the circuit. In other words, the current in the circuit is related to the difference of the two emf's: one is the (steady) emf of battery ε and the other is the (changeable) emf self-induced εL in the inductor. This last one, has the formula ε_L = -L di/dt

where i is the current induced in the inductor and L is its inductivity. Over time, the rise in current due to the self-induced emf in the inductor becomes less rapid. As a result, the current in the circuit approaches the value of the steady current ε calculated through the Ohm's law. However, in presence of an inductor in the circuit, the current never reaches the ε/R value (or using the language of mathematics, we say the current reaches this limit value in an infinite time interval). Therefore, we say: "An inductor initially opposes the rise in the current in the circuit but after a long time, it acts as a simple conducting wire." Thus, when the switch is at position a as discussed earlier, the original circuit behaves like the simplified one shown below: Since numerically the value of the self-induced emf in the best case can equal the value of emf produced by the battery, the current i in the circuit is in the direction of the red arrow (here clockwise). Therefore, the potential in the resistor decreases in the clockwise direction and as a result, the potential difference across the resistor R is ∆V = -i ∙ R

Likewise, since the self-induced emf in the inductor is in the opposite direction to the emf of battery, we write again ε_L = -L di/dt

Hence, since the emf produced by the battery is clockwise, we write from the Kirchhoff's Second Law (the voltage law), which is based on the law of conservation of energy: ε+∆V+ε_i = 0

Or ε-i ∙ R-L ∙ di/dt = 0

We can rearrange the last equation to isolate ε: ε = i ∙ R+L ∙ di/dt

The solution of this differential equation in terms of the current i, (using the differentiation techniques which you can find in the math section of this webpage), is i(t) = ε/R ∙ (1-e^(-(R ∙ t)/L))

The form of this equation is similar to that of potential difference (and charge) in a RC circuit. If we write the term L/R as τL (we call it "the inductive time constant), the above equation is written as i(t) = ε/R ∙ (1-e^(-t/τ_L ))

This equation is used to calculate the current at any instant when the current in the circuit is rising. When the current drops, we use the equation i(t) = ε/R ∙ e^(-t/τ_L )

to calculate the current in the circuit at any instant t. Example: A 20 Ω resistor is connected to a 12V battery. A 16 cm long inductor having 4000 turns and the area of each loop equal to 8 cm2, is connected in series to the resistor, as shown in the figure. Calculate: a) The current in the circuit when the switch S is in the actual position (OFF) b) The current in the circuit immediately after the switch is closed (ON) c) The current in the circuit 0.003 s after the switch is closed (ON) d) The current in the circuit 10 s after the switch is closed (ON) e) Can you find the exact time when the current becomes steady? Why? f) What is the potential difference in the resistor and inductor at t = 0.001 s? Solution a) The circuit is initially open (OFF). Hence, no current is flowing through the circuit when the switch is in this position. b) We have t = 0 immediately after the switch turns ON. In this case, we use the equation i(t) = ε/R ∙ (1-e^(-(R ∙ t)/L))

to find the current in the circuit at this instant. Thus, we obtain i(t) = ε/R ∙ (1-e^(-(R ∙ t)/L) ) = 12/20 ∙ (1-e^(-(20 ∙ 0)/L) ) = 12/20 ∙ (1-e^0 ) = 12/20 ∙ (1-1) = 12/20 ∙ 0 = 0

Therefore, the current is still zero immediately after the switch turns ON, as the self-induced emf in the coil prevents the current from flowing in the circuit. c) Before calculating the current in other instants, we must find the self-inductance L in the inductor using the equation L = (μ_{0} ∙ N^{2} ∙ A)/l

where N = 4000 = 4 × 103 turns A = 8 cm2 = 8 × 10-4 m2 l = 16 cm = 0.16 m (μ0 = 4π × 10-7 N/A2) Thus, we obtain for the self-inductance L of the inductor: L = ((4 ∙ 3.14 ∙ 10^{-7} N/A^{2} ) ∙ (4 × 10^{3} )^{2} ∙ (8 × 10^{-4} m^{2} ))/((0.16 m) ) = 10048 × 10^{-5} H = 0.1 H

The current flowing in the circuit at t = 0.003 s is: i(t) = ε/R ∙ (1-e^(-(R ∙ t)/L) ) = 12/20 ∙ (1-e^(-(20 ∙ 0.001)/0.1) ) = 0.6 A ∙ (1-e^(-0.2) ) = 0.6 ∙ (1-0.819) = 0.6 A ∙ 0.181 = 0.109 A

This means the current is rising but it has not reached the maximum value (0.6 A) yet. d) Now, we find the current in the circuit for t = 10s. We have i(t) = ε/R ∙ (1-e^(-(R ∙ t)/L) ) = 12/20 ∙ (1-e^(-(20 ∙ 10)/0.1) ) = 0.6 ∙ (1-e^(-2000) ) = 0.6 ∙ (1-0) = 0.6 ∙ 1 = 0.6 A

This results means the current in the circuit is already steady. e) It is not possible to find an exact time in which the current reaches the maximum value (0.6 A) and becomes steady because the current rises in an asymptotic fashion, i.e. initially the current rises faster and then the rise slows down more and more, until it becomes undetectable. f) We use the value found for the current in (c) to calculate the potential difference in the resistor and inductor in the instant t = 0.001 s. Thus, ∆V_((resistor) ) = i(t) ∙ R = (0.109 A) ∙ (20 Ω) = 2.18 V

and ∆V_((inductor) ) = ε-∆V_((resistor) ) = 12 V-2.18 V = 9.82 V

What is the Dimension of τ? In the previous paragraph, we gave the definition of the constant τ as the ratio of inductance L of the coil and resistance R of the resistor, i.e. τ = L/R

Let's express the units of inductance and resistance in terms of the fundamental SI units and see what it gives. Thus, since L = (μ_{0} ∙ N^{2} ∙ A)/l

we have for the unit of inductance L: [Unit of L] = ([N/A^{2} ] ∙ [m^{2} ])/[m]

Giving that 1N = 1 (kg ∙ m)/s^{2}

we obtain [Unit of L] = ([((kg ∙ m)/s^{2} )/A^{2} ] ∙ [m^{2} ])/[m] = [(kg ∙ m^{2})/(A^{2} ∙ s^{2} )]

As for the unit of resistance, we have from the Ohm's Law [Unit of resistance] = [(Unit of emf)/(Unit of current)] = [V/A]

By definition, emf is the work done to move the charges throughout the circuit. Thus, 1 [V] = 1[J/C]

Hence, since 1 C = 1 A ∙ s, we obtain [Unit of resistance] = [(J/C)/A] = [J/(A ∙ C)] = [(N ∙ m)/(A ∙ A ∙ s)] = [((kg ∙ m)/s^{2} ∙ m)/(A^{2} ∙ s)] = [(kg ∙ m^{2})/(A^{2} ∙ s^{3} )]

Therefore, we obtain for the unit of τ: [Unit of τ] = [(Unit of L)/(Unit of R)] = [((kg ∙ m^{2})/(A^{2} ∙ s^{2} ))/((kg ∙ m^{2})/(A^{2} ∙ s^{3} ))] = [s]

Therefore, the unit of inductive time constant τ is the same as the unit of time, second [s] as expected (giving that it is a time constant). This is obvious, since the value of exponential part of the equation i(t) = ε/R ∙ (1-e^(-(R ∙ t)/L) )

must be dimensionless, in order to subtract it from 1. 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) = i_{0} ∙ e^(- t/τ_L )

where τL = 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: 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: a) The initial value of current in the circuit b) The value of current in the circuit 0.02s after the switch turns OFF c) The potential differences across the resistor and inductor 0.02 s after the switch turns OFF d) The value of current in the circuit 20 s after the switch turns OFF. Solution a) 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 τL = L/R, we obtain i(0) = (48/12 A) ∙ e^(-(12 ∙ 0)/5) = (4A) ∙ e^0 = (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

b) 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. c) 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

d) Since τL = 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: 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 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 imax = 0.8 ∙ imax 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 imax from both sides, we obtain 0.8 = ├ 1-e^(-10t) ┤ e^(-10t) = 1-0.8 e^(-10t) = 0.2 ln(e^(-10t) ) = ln0.2 -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. Summary A RL circuit contains at least one resistor and one solenoid along with other useful components. In other words, in RL circuits, at least one resistor and one inductor are connected in the same wire. Inductor is one of the major passive components in electronics. The basic passive components in electronics are resistors, capacitors, inductors and transformers. RL circuits are similar in concept to RC circuits. However, capacitors and inductors have different construction properties, limitations and usage. A rise or a fall of the current through a RL circuit occurs when a source by an electromotive force ε supplies a single loop circuit containing a resistor R and an inductor L. From the Kirchhoff's Second Law (the voltage law), which is based on the law of conservation of energy, we have: ε+∆V+ε_i = 0

Or ε-i ∙ R-L ∙ di/dt = 0

The rise in current in a RL circuit as a function of time t, is: i(t) = ε/R ∙ (1-e^(-t/τ_L ))

while the current fall in such circuits after the switch turns OFF, is i(t) = ε/R ∙ e^(-t/τ_L )

where τ_L = L/R

is known as the inductive time constant. It has the unit of time (second). The following rule is applied in the RL circuits: "An inductor initially opposes the rise in the current in the circuit but after a long time, it acts as a simple conducting wire." We can apply a logarithmic approach to calculate the time in which the current flowing in the circuit reaches a certain part of the initial or maximum value of current. **1)** A 4Ω resistor and a 2H inductor are connected in the same circuit supplied by a 20V battery. What is the current in the circuit 0.01 s after the switch is moved at the position a?

- 5.0 A
- 4.9 A
- 0.2 A
- 0.1 A

**Correct Answer: D**

**2)** A RL circuit having a 10 Ω resistor and a 0.2 H inductor connected in series, is supplied by a 24 V battery, as shown in the figure.

What is the current flowing through the circuit 0.05 s after the switch turns OFF after being ON for a long time?

- 0
- 0.197 A
- 1.97 A
- 2.4 A

**Correct Answer: B**

**3)** A RC circuit that contains a 18Ω resistance and a 2H inductor is supplied by a battery, as shown in the figure.

How long does it take the current to drop at 20% of its initial value after the switch S turns OFF?

- 0.077 s
- 1.61 s
- 0.161 s
- 0.179 s

**Correct Answer: D**

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