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By the end of the section, you will be able to:
  • Describe how the current varies in a resistor, a capacitor, and an inductor while in series with an ac power source
  • Use phasors to understand the phase angle of a resistor, capacitor, and inductor ac circuit and to understand what that phase angle means
  • Calculate the impedance of a circuit

The ac circuit shown in [link] , called an RLC series circuit , is a series combination of a resistor, capacitor, and inductor connected across an ac source. It produces an emf of

v ( t ) = V 0 sin ω t .
Figure a shows a circuit with an AC voltage source connected to a resistor, a capacitor and an inductor in series. The source is labeled V0 sine omega t. Figure b shows sine waves of AC voltage and current on the same graph. Voltage has a greater amplitude than current and its maximum value is marked V0 on the y axis. The maximum value of current is marked I0. The two curves have the same wavelength but are out of phase. The voltage curve is labeled V parentheses t parentheses equal to V0 sine omega t. The current curve is labeled I parentheses t parentheses equal to I0 sine parentheses omega t minus phi parentheses.
(a) An RLC series circuit. (b) A comparison of the generator output voltage and the current. The value of the phase difference ϕ depends on the values of R , C , and L .

Since the elements are in series, the same current flows through each element at all points in time. The relative phase between the current and the emf is not obvious when all three elements are present. Consequently, we represent the current by the general expression

i ( t ) = I 0 sin ( ω t ϕ ) ,

where I 0 is the current amplitude and ϕ is the phase angle    between the current and the applied voltage. The phase angle is thus the amount by which the voltage and current are out of phase with each other in a circuit. Our task is to find I 0 and ϕ .

A phasor diagram involving i ( t ) , v R ( t ) , v C ( t ) , and v L ( t ) is helpful for analyzing the circuit. As shown in [link] , the phasor representing v R ( t ) points in the same direction as the phasor for i ( t ) ; its amplitude is V R = I 0 R . The v C ( t ) phasor lags the i ( t ) phasor by π / 2 rad and has the amplitude V C = I 0 X C . The phasor for v L ( t ) leads the i ( t ) phasor by π / 2 rad and has the amplitude V L = I 0 X L .

Figure shows the coordinate axes, with four arrows starting from the origin. Arrow V subscripts R points up and right, making an angle omega t minus phi with the x axis. Its y intercept is V subscript R parentheses t parentheses. Arrow I0 is along arrow V subscript R, but shorter than it. Arrow V subscript L points up and left and is perpendicular to V subscript R. It makes a y intercept V subscript L parentheses t parentheses. Arrow V subscript C points down and right. It is perpendicular to V subscript R. It makes a y intercept V subscript C parentheses t parentheses. Three arrows labeled omega are each perpendicular to V subscript R, V subscript L and V subscript C, shown near their tips.
The phasor diagram for the RLC series circuit of [link] .

At any instant, the voltage across the RLC combination is v R ( t ) + v L ( t ) + v C ( t ) = v ( t ) , the emf of the source. Since a component of a sum of vectors is the sum of the components of the individual vectors—for example, ( A + B ) y = A y + B y —the projection of the vector sum of phasors onto the vertical axis is the sum of the vertical projections of the individual phasors. Hence, if we add vectorially the phasors representing v R ( t ) , v L ( t ) , and v C ( t ) and then find the projection of the resultant onto the vertical axis, we obtain

v R ( t ) + v L ( t ) + v C ( t ) = v ( t ) = V 0 sin ω t .

The vector sum of the phasors is shown in [link] . The resultant phasor has an amplitude V 0 and is directed at an angle ϕ with respect to the v R ( t ) , or i ( t ), phasor. The projection of this resultant phasor onto the vertical axis is v ( t ) = V 0 sin ω t . We can easily determine the unknown quantities I 0 and ϕ from the geometry of the phasor diagram. For the phase angle,

ϕ = tan −1 V L V C V R = tan −1 I 0 X L I 0 X C I 0 R ,

and after cancellation of I 0 , this becomes

ϕ = tan −1 X L X C R .

Furthermore, from the Pythagorean theorem,

V 0 = V R 2 + ( V L V C ) 2 = ( I 0 R ) 2 + ( I 0 X L I 0 X C ) 2 = I 0 R 2 + ( X L X C ) 2 .
Three arrows start from the origin on the coordinate axis. Arrow V subscript R points up and right, making an angle omega t minus phi with the x axis. Arrow V0 points up and right, making an angle omega t with the x axis. It makes an angle phi with the arrow V subscript R. It makes a y intercept labeled V0 sine omega t. The third arrow is labeled V subscript L minus V subscript C. It points up and left and is perpendicular to arrow V subscript R. Dotted lines indicate that the rectangle formed with its longer side being V subscript R and shorter side being V subscript L minus V subscript C, would have the arrow V0 as a diagonal. An arrow labeled omega is shown near the tip of V subscript R, perpendicular to it.
The resultant of the phasors for v L ( t ) , v C ( t ) , and v R ( t ) is equal to the phasor for v ( t ) = V 0 sin ω t . The i ( t ) phasor (not shown) is aligned with the v R ( t ) phasor.

The current amplitude is therefore the ac version of Ohm’s law:

I 0 = V 0 R 2 + ( X L X C ) 2 = V 0 Z ,

where

Z = R 2 + ( X L X C ) 2

is known as the impedance    of the circuit. Its unit is the ohm, and it is the ac analog to resistance in a dc circuit, which measures the combined effect of resistance, capacitive reactance, and inductive reactance ( [link] ).

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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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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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
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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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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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