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15.3 Rlc series circuits with ac  (Page 2/4)

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Photograph of power capacitors at a power station.
Power capacitors are used to balance the impedance of the effective inductance in transmission lines.

The RLC circuit is analogous to the wheel of a car driven over a corrugated road ( [link] ). The regularly spaced bumps in the road drive the wheel up and down; in the same way, a voltage source increases and decreases. The shock absorber acts like the resistance of the RLC circuit, damping and limiting the amplitude of the oscillation. Energy within the wheel system goes back and forth between kinetic and potential energy stored in the car spring, analogous to the shift between a maximum current, with energy stored in an inductor, and no current, with energy stored in the electric field of a capacitor. The amplitude of the wheel’s motion is at a maximum if the bumps in the road are hit at the resonant frequency, which we describe in more detail in Resonance in an AC Circuit .

Figure shows one wheel of a car. Arrows show the up-down motion of its shock absorber spring.
On a car, the shock absorber damps motion and dissipates energy. This is much like the resistance in an RLC circuit. The mass and spring determine the resonant frequency.

Problem-solving strategy: ac circuits

To analyze an ac circuit containing resistors, capacitors, and inductors, it is helpful to think of each device’s reactance and find the equivalent reactance using the rules we used for equivalent resistance in the past. Phasors are a great method to determine whether the emf of the circuit has positive or negative phase (namely, leads or lags other values). A mnemonic device of “ELI the ICE man” is sometimes used to remember that the emf (E) leads the current (I) in an inductor (L) and the current (I) leads the emf (E) in a capacitor (C).

Use the following steps to determine the emf of the circuit by phasors:

  1. Draw the phasors for voltage across each device: resistor, capacitor, and inductor, including the phase angle in the circuit.
  2. If there is both a capacitor and an inductor, find the net voltage from these two phasors, since they are antiparallel.
  3. Find the equivalent phasor from the phasor in step 2 and the resistor’s phasor using trigonometry or components of the phasors. The equivalent phasor found is the emf of the circuit.

An RLC Series circuit

The output of an ac generator connected to an RLC series combination has a frequency of 200 Hz and an amplitude of 0.100 V. If R = 4.00 Ω , L = 3.00 × 10 −3 H, and C = 8.00 × 10 −4 F, what are (a) the capacitive reactance, (b) the inductive reactance, (c) the impedance, (d) the current amplitude, and (e) the phase difference between the current and the emf of the generator?

Strategy

The reactances and impedance in (a)–(c) are found by substitutions into [link] , [link] , and [link] , respectively. The current amplitude is calculated from the peak voltage and the impedance. The phase difference between the current and the emf is calculated by the inverse tangent of the difference between the reactances divided by the resistance.

Solution

  1. From [link] , the capacitive reactance is
    X C = 1 ω C = 1 2 π ( 200 Hz ) ( 8.00 × 10 −4 F ) = 0.995 Ω .
  2. From [link] , the inductive reactance is
    X L = ω L = 2 π ( 200 Hz ) ( 3.00 × 10 −3 H ) = 3.77 Ω .
  3. Substituting the values of R , X C , and X L into [link] , we obtain for the impedance
    Z = ( 4.00 Ω ) 2 + ( 3.77 Ω − 0.995 Ω ) 2 = 4.87 Ω .
  4. The current amplitude is
    I 0 = V 0 Z = 0.100 V 4.87 Ω = 2.05 × 10 −2 A .
  5. From [link] , the phase difference between the current and the emf is
    ϕ = tan −1 X L − X C R = tan −1 2.77 Ω 4.00 Ω = 0.607 rad .

Significance

The phase angle is positive because the reactance of the inductor is larger than the reactance of the capacitor.

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Check Your Understanding Find the voltages across the resistor, the capacitor, and the inductor in the circuit of [link] using v ( t ) = V 0 sin ω t as the output of the ac generator.

v R = ( V 0 R / Z ) sin ( ω t − ϕ ) ; v C = ( V 0 X C / Z ) sin ( ω t − ϕ + π / 2 ) = − ( V 0 X C / Z ) cos ( ω t − ϕ ) ; v L = ( V 0 X L / Z ) sin ( ω t − ϕ + π / 2 ) = ( V 0 X L / Z ) cos ( ω t − ϕ )

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Summary

  • An RLC series circuit is a resistor, capacitor, and inductor series combination across an ac source.
  • The same current flows through each element of an RLC series circuit at all points in time.
  • The counterpart of resistance in a dc circuit is impedance, which measures the combined effect of resistors, capacitors, and inductors. The maximum current is defined by the ac version of Ohm’s law.
  • Impedance has units of ohms and is found using the resistance, the capacitive reactance, and the inductive reactance.

Conceptual questions

In an RLC series circuit, can the voltage measured across the capacitor be greater than the voltage of the source? Answer the same question for the voltage across the inductor.

yes for both

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Problems

What is the impedance of a series combination of a 50 -Ω resistor, a 5.0 - μ F capacitor, and a 10 - μ F capacitor at a frequency of 2.0 kHz?

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A resistor and capacitor are connected in series across an ac generator. The emf of the generator is given by v ( t ) = V 0 cos ω t , where V 0 = 120 V, ω = 120 π rad/s , R = 400 Ω , and C = 4.0 μ F . (a) What is the impedance of the circuit? (b) What is the amplitude of the current through the resistor? (c) Write an expression for the current through the resistor. (d) Write expressions representing the voltages across the resistor and across the capacitor.

a. 770 Ω ; b. 0.16 A; c. I = ( 0.16 A ) cos ( 120 π t ) ; d. v R = 120 cos ( 120 π t ) ; v C = 120 cos ( 120 π t − π / 2 )

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A resistor and inductor are connected in series across an ac generator. The emf of the generator is given by v ( t ) = V 0 cos ω t , where V 0 = 120 V and ω = 120 π rad/s; also, R = 400 Ω and L = 1.5 H . (a) What is the impedance of the circuit? (b) What is the amplitude of the current through the resistor? (c) Write an expression for the current through the resistor. (d) Write expressions representing the voltages across the resistor and across the inductor.

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In an RLC series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively, an R = 500 Ω , L = 0.20 H, and C = 2.0 μ F . (a) What is the impedance of the circuit? (b) What is the amplitude of the current from the source? (c) If the emf of the source is given by v ( t ) = ( 100 V ) sin 1000 π t , how does the current vary with time? (d) Repeat the calculations with C changed to 0.20 μ F .

a. 690 Ω ; b. 0.15 A; c. I = ( 0.15 A ) sin ( 1000 π t − 0.753 ) ; d. 1100 Ω , 0.092 A, I = ( 0.092 A ) sin ( 1000 π t + 1.09 )

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An RLC series circuit with R = 600 Ω , L = 30 mH, and C = 0.050 μ F is driven by an ac source whose frequency and voltage amplitude are 500 Hz and 50 V, respectively. (a) What is the impedance of the circuit? (b) What is the amplitude of the current in the circuit? (c) What is the phase angle between the emf of the source and the current?

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For the circuit shown below, what are (a) the total impedance and (b) the phase angle between the current and the emf? (c) Write an expression for i ( t ) .

Figure shows a circuit with a voltage source 170 V, sine 120 pi t, a resistor of 5 ohm, a capacitor of 400 microfarad and an inductor of 25 milihenry all connected in series.

a. 5.7 Ω ; b. 29 ° ; c. I ​ = ( 30. A ) cos ( 120 π t )

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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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