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By the end of this section, you will be able to:
  • Analyze circuits that have an inductor and resistor in series
  • Describe how current and voltage exponentially grow or decay based on the initial conditions

A circuit with resistance and self-inductance is known as an RL circuit. [link] (a) shows an RL circuit consisting of a resistor, an inductor, a constant source of emf, and switches S 1 and S 2 . When S 1 is closed, the circuit is equivalent to a single-loop circuit consisting of a resistor and an inductor connected across a source of emf ( [link] (b)). When S 1 is opened and S 2 is closed, the circuit becomes a single-loop circuit with only a resistor and an inductor ( [link] (c)).

Figure a shows a resistor R and an inductor L connected in series with two switches which are parallel to each other. Both switches are currently open. Closing switch S1 would connect R and L in series with a battery, whose positive terminal is towards L. Closing switch S2 would form a closed loop of R and L, without the battery. Figure b shows a closed circuit with R, L and the battery in series. The side of L towards the battery, is at positive potential. Current flows from the positive end of L, through it, to the negative end. Figure c shows R and L connected in series. The potential across L is reversed, but the current flows in the same direction as in figure b.
(a) An RL circuit with switches S 1 and S 2 . (b) The equivalent circuit with S 1 closed and S 2 open. (c) The equivalent circuit after S 1 is opened and S 2 is closed.

We first consider the RL circuit of [link] (b). Once S 1 is closed and S 2 is open, the source of emf produces a current in the circuit. If there were no self-inductance in the circuit, the current would rise immediately to a steady value of ε / R . However, from Faraday’s law, the increasing current produces an emf V L = L ( d I / d t ) across the inductor. In accordance with Lenz’s law, the induced emf counteracts the increase in the current and is directed as shown in the figure. As a result, I(t) starts at zero and increases asymptotically to its final value.

Applying Kirchhoff’s loop rule to this circuit, we obtain

ε L d I d t I R = 0 ,

which is a first-order differential equation for I(t) . Notice its similarity to the equation for a capacitor and resistor in series (See RC Circuits ). Similarly, the solution to [link] can be found by making substitutions in the equations relating the capacitor to the inductor. This gives

I ( t ) = ε R ( 1 e R t / L ) = ε R ( 1 e t / τ L ) ,

where

τ L = L / R

is the inductive time constant    of the circuit.

The current I(t) is plotted in [link] (a). It starts at zero, and as t , I(t) approaches ε / R asymptotically. The induced emf V L ( t ) is directly proportional to dI / dt , or the slope of the curve. Hence, while at its greatest immediately after the switches are thrown, the induced emf decreases to zero with time as the current approaches its final value of ε / R . The circuit then becomes equivalent to a resistor connected across a source of emf.

Figure a shows the graph of electric current I versus time t. Current increases with time in a curve which flattens out at epsilon I R. At t equal to tau subscript L, the value of I is 0.63 epsilon I R. Figure b shows the graph of magnitude of induced voltage, mod V subscript L, versus time t. Mod V subscript L starts at value epsilon and decreases with time till the curve reaches zero. At t equal to tau subscript L, the value of I is 0.37 epsilon.
Time variation of (a) the electric current and (b) the magnitude of the induced voltage across the coil in the circuit of [link] (b).

The energy stored in the magnetic field of an inductor is

U L = 1 2 L I 2 .

Thus, as the current approaches the maximum current ε / R , the stored energy in the inductor increases from zero and asymptotically approaches a maximum of L ( ε / R ) 2 / 2 .

The time constant τ L tells us how rapidly the current increases to its final value. At t = τ L , the current in the circuit is, from [link] ,

I ( τ L ) = ε R ( 1 e −1 ) = 0.63 ε R ,

which is 63 % of the final value ε / R . The smaller the inductive time constant τ L = L / R , the more rapidly the current approaches ε / R .

We can find the time dependence of the induced voltage across the inductor in this circuit by using V L ( t ) = L ( d I / d t ) and [link] :

V L ( t ) = L d I d t = ε e t / τ L .

The magnitude of this function is plotted in [link] (b). The greatest value of L ( d I / d t ) is ε ; it occurs when dI/dt is greatest, which is immediately after S 1 is closed and S 2 is opened. In the approach to steady state, dI/dt decreases to zero. As a result, the voltage across the inductor also vanishes as t .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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