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5. Application of Ordinary Differential Equations: Series RL Circuit

RL circuit diagram

The RL circuit shown has a resistor and an inductor connected in series. A constant voltage V is applied when the switch is closed.

The (variable) voltage across the resistor is given by:

The (variable) voltage across the inductor is given by:

Kirchhoff's voltage law says that the directed sum of the voltages around a circuit must be zero. This results in the following differential equation:

Once the switch is closed, the current in the circuit is not constant. Instead, it will build up from zero to some steady state.

Solving the DE for a Series RL Circuit

The solution of this differential equation is:

Proof

[We did the same problem but with particular values back in section 2. Separation of Variables]

Here is the graph of this equation:

The plot shows the transition period during which the current adjusts from its initial value of zero to the final value V/R, which is the steady state.

The Time Constant

The time constant (TC), known as τ, of the function

is the time at which R/L is unity. Thus for the RL transient, the time constant τ = L/R seconds.

NOTE: τ is the Greek letter "tau" and is not the same as the time variable t.

At 1 τ

At this time the current is 63.2% of its final value.

Similarly at 2 τ,

1 - e^-2 = 1 - 0.135 = 0.865

The current is 86.5% of its final value.

After 5 τ the transient is generally regarded as terminated. For convenience, the time constant τ is the unit used to plot the current of the equation

That is, since τ = L/R, we think of it as:

---

Let's now look at some examples of RL circuits.

EXAMPLE 1

An RL circuit has an emf of 5 V, a resistance of 50 W, an inductance of 1 H, and no initial current.

Find the current in the circuit at any time t. Distinguish between the transient and steady-state current.

Answer

EXAMPLE 2

A series RL circuit with R = 50 W and L = 10 H has a constant voltage V = 100 V applied at t = 0 by the closing of a switch.

Find

(a) the equation for i (you may use the formula rather than DE),

(b) the current at t = 0.5 s

(c) the expressions for V_R and V_L

(d) the time at which V_R = V_L

Answer

Two-mesh Circuits

The next two examples are "two-mesh" types where the differential equations become more sophisticated. We will use Scientific Notebook to do the grunt work once we have set up the correct equations.

EXAMPLE 3

In the two-mesh network shown below, the switch is closed at t = 0 and the voltage source is given by V = 150 sin 1000t V. Find the mesh currents i₁ and i₂ as given in the diagram.

Answer

EXAMPLE 4

The switch is closed at t = 0 in the two-mesh network shown below. The voltage source is given by V = 30 sin 100t V. Find the mesh currents i₁ and i₂ as given in the diagram.

Answer

 

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Chapter Contents

• Differential Equations
• Predicting AIDS - a DEs example
• 1. Solving Differential Equations
• 2. Separation of Variables
• 3. Integrable Combinations
• 4. Linear DEs of Order 1
• 5. Application: RL Circuits
• 6. Application: RC Circuits
• 7. Second Order DEs - Homogeneous
• 8. Second Order DEs - Damping - RLC
• 9. Second Order DEs - Forced Response
• 10. Second Order DEs - Solve Using SNB

Following are the original SNB files (.tex or .rap) used in making this chapter. For more information, go to SNB info.

SNB files

Predicting AIDS - a DEs example

7. Second Order DEs - Homogeneous

8. Second Order DEs - Damping - RLC

9. Second Order DEs - Forced Response

10. Second Order DEs - Solve Using SNB

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