rl and rc circuit

7/29/2019 Rl and Rc Circuit

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Copyright Rene Barrientos Page 1

RL CIRCUITAND RC CIRCUITS

The RL Circuit

An RL circuit is one which contains aresistor and an inductor. Resistors are the load placed in a circuit and some

examples are a light-bulb or a hair dryer. An inductor is essentially a coil (solenoid) whose function is to impede

changes in current.

The science behind these electrical components is covered in a second semester in introductory physics. For ourpurposes, we just need to know that charge moves because it is acted upon by an electric field created by the power

source (e.g. a battery or electric outlet). As the circuit is transverse in the direction of the current, work is done by

the field at each component and manifests itself as a voltage drop. In the case of inductors, a changing current

gives rise to self-inductance. The mathematics of this phenomenon is described byFaradays Law1

.

where is the magnetic flux induced by the changing current in the coil. Consider then the circuit shown belowcomposed of a power source (say a battery delivering voltage V), a resistor of resistance R (ohms) and an inductor

with inductance L (henrys):

When the switch is closed, Faradays Law tell us that

Rewriting this differential equation in standards form,

and we may assume that 0 0.This equation can be solved by separation of variables (if is constant), by using an integrating factor, or evenusing the Laplace transform. Using an integrating factor, for example, gives the following result:

; /

Thus,

/

Integrating from 0 to t: 0 /

Since

0 0,

/

/ 1

1 Michael Faraday (1791-1867) was an English physicist and chemist who contributed substantially to the fields ofelectromagnetic and electrochemistry. He established the basis for electromagnetic field concept, which was central to the workof James Clerk Maxwell.

V

R

L

i

+


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Solving for /

As expected, when the current builds up, self-induction ceases to play a role and the solenoid becomes just another

conducting portion in the circuit. Then, the current is given by Ohms Law2:

; Example 1 Find the current at / for the circuit shown below. Assume 0 0.

Solution

Substituting the values in the solution for: 124 1

31 When /, 1 . Thus,

31 1 3 0.63That is, at / is 63% of its steady state value.

The RC Circu it

An RC circuit is composed of a power supply, a load (resistor) and acapacitor. A capacitor consists of two parallel

plates and is represented by the symbol

and we use the letter C (forcapacitance) to identify them in circuits. Its unit is the farad (but more often themicrofarad).

When a current flows in the circuit, the capacitor stores charge in its plates. If the circuit is interrupted, the capacitor

discharges thus maintaining a current for a some time (until it is fully discharged). Let be the charge in thecapacitor at time t. Then the magnitude of voltage drop across a capacitor is given by

1

where is the charge in the capacitor at time .

2Georg Simon Ohm (1789 1854) was a German physicist who worked on the theory of electricity.

12volts

4

0.2 hi

+

R

C+

V


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The differential equation governs the current and charge in a circuit can be obtained from Kirchhoffs Law3:

1 0This equation involves both the current and the charge in the capacitor. We can correct this by noting thatcurrent is flow of charge per unit time:

Thus,

1 0

This may be written as

;

Let us first consider a simple example in which a circuit, originally opened, has an initial charge in its capacitor and

contains no source of emf (no battery). When it is closed, the capacitor discharges and a current will appear in the

circuit. Let us find an expression for that current.

The differential equation for this case is a linear, homogeneous equation with constant coefficients:

1 0; 0 The characteristic equation is

1 0whose roots are

1/Thus, / is the general solution. Applying the initial condition 0 ,

/Therefore, the current decays exponentially and has a value of 1/ at . Notice that 0 as .where did the charge that was in the capacitor go? Research this question.

Example 2 A circuit is hooked to a 30 volt battery and has a 1/120

resistor, and a 12 f capacitor. Assume0 400 coul.Solution

The equation governing this system is

1120

11210 30; 0 400 10 coul

or 10

7 360; 0 400 106 coulAn integrating factor fir the last equation is . Thus,

10

7

360107

; 0 400 106 coulIntegrating, 0 360

3Gustav Robert Kirchhoff (1824 1887) was a German physicist. His Laws of Electric Circuits state that the algebraic sum of allvoltage drops (potential differences) along a closed loop in a circuit is 0 and that the algebraic sum of currents at a node is zero.


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400 10 360

400 10 360 101 0.0004 0.0000361

Since

/,

0.00003610 360

The current decrease to 0 very quickly. Notice that the terminal charge in capacitor is 0.0004 0.000036or436 coul so the current added to the charge already stored in the capacitor.

Exercise Solve examples 1 and 2 using the Laplace transform

5. 10-8

1.10-7

1.5 10-7

2. 10-7

2.5 10-7

3. 10-7

0.000405

0.000410

0.000415

0.000420

0.000425

0.000430

0.000435

5.10-8

1.10-7

1.5 10-7

2.10-7

.5 10-7

3.10-7

50

100

150

200

250

300

350

The charge in the capacitor increases untilit reaches a terminal value of436 coul

The current in the circuit decrease until itreaches a terminal value of0 amps.

rl and rc circuit

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