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RL Circuits. AP Physics C Montwood High School R. Casao. A circuit that contains a coil, such as a solenoid, has a self-inductance that prevents the current from increasing or decreasing instantaneously. A circuit element that has a large inductance is called an inductor.
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RL Circuits AP Physics C Montwood High School R. Casao
A circuit that contains a coil, such as a solenoid, has a self-inductance that prevents the current from increasing or decreasing instantaneously. • A circuit element that has a large inductance is called an inductor. • The circuit symbol for an inductor is • We will always assume that the self-inductance of the remainder of the circuit is negligible compared to that of the inductance. • Keep in mind that even a circuit without a coil has some self-inductance that can affect the behavior of the circuit.
Because the inductance of the inductor esults in a back EMF, an inductor in a circuit opposes changes in the current in that circuit. • The inductor attempts to keep the current the same as it was before the change occurred. • If the battery voltage in the circuit is increased so that the current rises, the inductor opposes this change and the rise is not instantaneous. • If the battery voltage is decreased, the presence of the inductor results in a slow drop in the current rather than an immediate drop. • The inductor causes the circuit to be “sluggish” as it reacts to changes in the voltage.
Consider a circuit consisting of a resistor, an inductor, and a battery. • The internal resistance of the battery will be neglected. • The switch is closed at t = 0 and the current begins to increase. • Due to the increasing current, the inductor will produce an EMF (the back EMF) that opposes the increasing current.
The inductor acts like a battery whose polarity is opposite that of the battery in the circuit. • The back EMF produced by the inductor is: • Since the current is increasing, dI/dt is positive; therefore, EMFL is negative.
This corresponds to the fact that there is a potential drop in going from a to b across the inductor. • Point a is at a higher potential than point b. • Apply Kirchhoff’s loop rule to the circuit:
Divide through by R: • Let ; EMF and R are constant; I is variable • Take the derivative of x:
Replace the variables in the equation: • Combine like terms and differentiate the differential equation:
Exponentiate both sides of the equation: • At t = 0 s, I = 0 A, so:
Replacing the substituted variables: • The current I as a function of time:
The equation shows how the inductor effects the current. • The current does not increase instantly to its final equilibrium value when the switch is closed but instead increases according to an exponential function. • If the inductance approaches zero, the exponential term becomes zero and there is no time dependence for the current – the current increases instantaneously to its final equilibrium value in the absence of the inductance. • Time constant for an RL circuit:
The current I as a function of time: • The time constant τ is the time it takes for the current to reach (1 – e-1) = 0.63 of the final value, EMF/R. • The graph of current versus time, where I = 0 A at t = 0 s; the final equilibrium value of the current occurs at t = and is equal to EMF/R. • The current rises very fast initially and then gradually approaches the equilibrium value EMF/R as t .
The rate of increase of current dI/dt is a maximum (equal to EMF/L) at t = 0 s and decreases exponentially to zero as t . • Consider the RL circuit shown in the figure below:
The curved lines on the switch S represent a switch that is connected either to a or b at all times. If the switch is connected to neither a nor b, the current in the circuit suddenly stops. • Suppose that the switch has been set at position a long enough to allow the current to reach its equilibrium value EMF/R. The circuit is described by the outer loop only. • If the switch is thrown from a to b, the circuit is described by just the right hand loop and we have a circuit with no battery (EMF = 0 V).
Applying Kirchhoff’s loop rule to the right-hand loop at the instant the switch is thrown from a to b: • Divide through by R:
Combine the current I terms and differentiate the differential equation:
Exponentiate both sides of the equation: • The current I as a function of time for an RL circuit in which the current is decaying over time:
When t = 0 s, Io = EMF/R and τ = L/R: • The current is continually decreasing with time. • The slope, dI/dt, is always negative and has its maximum value at t = 0 s. • The negative slope indicates that the is now positive; that is, point a is at a lower potential than point b.
Time Constant of an RL Circuit • The circuit shown in the figure consists of a 30 mH inductor, a 6 resistor, and a 12 V battery. The switch is closed at t = 0 s. • Find the time constant of the circuit. • Determine the current in the circuit at t = 0.002 s.
Kirchhoff’s Rule and Inductors • For the circuit shown in the figure: • find the currents I1, I2, and I3 immediately after switch S is closed.
Answer: The current in an inductor cannot change instantaneously from 0 A to the steady state current, therefore, the current in the inductor must be zero after the switch is closed because it is 0 A before the switch is closed. • I3 = 0 A
The current in the left loop equals the EMF divided by the equivalent resistance of the two resistors in series. • Req = 10 + 20 = 30
find the currents I1, I2, and I3 a long time after switch S has been closed. • Answer: when the current reaches its steady-state value, dI/dt = 0, so there is no potential drop across the inductor. The inductor acts like a short circuit (a wire with zero resistance).
Parallel resistances: Total resistance = 10 + 10 = 20 Total current:
The 7.5 A current will split in half to travel through the two equal 20 resistors; I2 = I3 = 3.75 A • find the currents I1, I2, and I3 immediately after switch S has been opened.
Answer: The current in an inductor cannot change instantaneously from the steady state current to 0 A, therefore, the current in the inductor must be the same after the switch is opened as it was just before the switch is opened. • I3 = 3.75 A
When the switch is opened, I1 drops to 0 A. • To oppose the change in the direction of the magnetic flux through the inductor, the direction of the induced current in the loop changes direction; I2 = I3 = 3.75 A.
D. find the currents I1, I2, and I3 a long time after switch S has been opened. Answer: all currents must be 0 A a long time after the switch is opened. I1 = I2 = I3 = 0 A
RL Circuits Summary • The inductor acts like a wire when t is large: • The inductor acts like an open circuit when t = 0 s; i = 0 A • The current starts from 0 A and increases up to a maximum of with a time constant given by • As the current increases from 0 A to its steady-state value: • Current:
Voltage across the resistor: • Voltage across the inductor:
The potential difference across a resistor depends on the current. • The potential difference across an inductor depends on the rate of change of the current.
The potential difference across an inductor depends on the rate of change of the current.
VR (V) • When the EMF source is removed from the circuit and the current begins to decay from the steady-state current to 0 A: • Current decay: • Voltage across resistor: • Voltage across inductor:
What is Happening During Current Decay? • When the battery is removed and the RL series circuit is shorted, the current keeps flowing in the same direction it was for awhile. How can this be? • What is happening is that the current tries to drop suddenly, but this induces an EMF to oppose the change, causing the current to keep flowing for awhile. • Another way of thinking about it is that the magnetic field that was stored in the inductor is “collapsing.” • There is energy stored in the magnetic field, and when the source of current is removed, the energy flows from the magnetic field back into the circuit.
Make Before Break Switches • The switch in a circuit like the one at right has to be a special kind, called a “make before break” switch. • The switch has to make the connection to bbefore breaking the connection with a. • If the circuit is allowed to be in the state like this…even momentarily, midway between a and b, then a big problem results. • Recall that for a capacitor, when we disconnect the circuit the charge will merrily stay on the capacitor indefinitely. • Not so on an inductor. The inductor needs current, i.e. flowing charge. It CANNOT go immediately to zero. • The collapsing magnetic field in the inductor will force the current to flow, even when it has no where to go. • The current will flow in this case by jumping the air gap. Link to video You have probably seen this when unplugging something with a motor—a spark that jumps from the plug to the socket.
spark 12V Start Your Engines: The Ignition Coil • The gap between the spark plug in a combustion engine needs an electric field of ~107 V/m in order to ignite the air-fuel mixture. For a typical spark plug gap, one needs to generate a potential difference > 104 V. • But, the typical EMF of a car battery is 12 V. So, how does a spark plug work? • Breaking the circuit changes the current through the “primary coil” • Result: LARGE change in flux thru secondary -- large induced EMF • The “ignition coil” is a double layer solenoid: • Primary: small number of turns , connected to 12 V battery • Secondary: MANY turns -- connected to spark plug
