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Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13

PHYS 2420 AC Circuits Solutions

  1. Show that the RMS value of the pure AC part of a wave is given by

    2. .

    VDC is the DC offset of the wave V(t).

    Let V(t) be an function with both AC and DC components. The DC component is given by the average of V(t)

    .

    The RMS value of V(t) is defined

    .

    Then the purely AC part of the function is defined

    VAC(t) = V(t) - VDC .

    The RMS value of this function is then

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  2. The diagram below shows a triangular wave (often called a sawtooth wave). Over one cycle, the equation for the wave is

    3. Find the DC offset. Find the RMS value of the wave. Find the AC part of the wave. Find the RMS value of the pure AC part of the wave.

    The DC offset, or average value, of V(t) is

    This result is obvious if one recalls that the area of a triangle is half the base times the height.

    The RMS value of V(t) is

    Note that this value also does not depend on t1.

     

     The purely AC part of V(t) is

    Using the result of Question 1, we find that the RMS value of VAC(t) is

      

  3. The diagram below shows an exponentially-decaying wave. Over one cycle, the equation for the wave is for 0 £ t < T.
    1. Find the DC offset.
    2. Examine the DC offset in the limit that KT is large.
    3. Examine the DC offset in the limit that KT is small. The expansion will be of use.
    4. Find RMS value of the wave. Examine the limits of the RMS value. The expansion will be of use.
    5. Find the RMS value of the purely AC part of the wave. Examine the limits.

    6. The DC offset, or average value, of V(t) is

    7. For large kT, e-kT » 0. So VDC » V0/kT.
    8. Using the given expansion for small kT, we have

    9. The RMS value of the wave is

      10. For large kT, the exponential is approximately zero and .

      For small values of kT, we use the given expansion

      The last result occurs because (1-kT)1/2 » 1-1/2kT . As a result, we see that VRMS approaches the value of VDC in the limit of large kT.

    10. Using the formula from Question 1 for finding VAC-RMS, we have

      11. For the limit of large kT, the exponentials reduce to zero. We thus have

      .

      For small kT, we use MAPLE to handle the expansion and find

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  4. A single line carries two signals, as shown in the diagram below, V1 = 10.0sin(100t) and V2 = 7.0sin(10000t) where t is in seconds. A capacitor of 1.00 μF and a 1000-W resistor are connected in series. What is the voltage signal observed over the resistor?

    5. The first step to solving this problem is to realize that V1 and V2 may be treated separately. Let us consider V1 first. Let I1 be the current leaving the power supply.

    The capacitor and the resistor are in series so they share the same current, i.e. I1. For a capacitor the current leads the voltage by 90° . The voltage drop over the capacitor is related to I1 by VC1 = I1XC1, where XC1 = 1/Ω1C = 10,000 Ω. The current and voltage drop are in phase for the resistor and VR1 = I1R. The resulting phasor diagram is

    Using the Pythagorean Theorem we find

    Z1 = [(XC1)2 + R2]1/2 = [(10,000 Ω)2 + (1000 Ω)2]1/2 = 10,049.9 Ω.

    From Z1 we can now find the maximum or peak value of I1 using Ohm's Law

    I1max = V1max/Z1 = (10.0 Volts)/(10,049.9 Ω) = 0.000995 A.

    The phase angle is

    f1 = arctan-1(XC1/R) = arctan-1(10000/1000) = 84.29° .

    Note that the diagram shows that I1 leads V1 by φ1 or conversely that V1 lags I1. This indicates that the form of I1(t) is

    I1(t) = I1maxsin(Ω1t + φ1) = (0.000995 A)sin(100t + 84.29°) .

    Since VR1(t) and I1(t) are in phase, the voltage drop over the resistor can now be found using Ohm's Law

    VR1(t) = RI1(t) = (0.995 V) sin(100t + 84.29°) .

    Next we consider V2(t) and follow the same steps. Let I2 be the current leaving the power supply. Again, the capacitor and the resistor are in series so they share the same current, i.e. I2. For a capacitor the current leads the voltage by 90° . The voltage drop over the capacitor is related to I2 by VC2 = I2XC2, where XC2 = 1/Ω2C = 100 Ω. The current and voltage drop are in phase for the resistor and VR2 = I2R. The resulting phasor diagram is

    Using the Pythagorean Theorem we find

    Z2 = [(XC2)2 + R2]1/2 = [(100 Ω)2 + (1000 Ω)2]1/2 = 1004.99 Ω.

    From Z2 we can now find the maximum or peak value of I2 using Ohm's Law

    I2max = V2max/Z2 = (7.0 Volts)/(1004.99 Ω) = 0.00697 A.

    The phase angle is

    f2 = arctan(XC2/R) = arctan(100/1000) = 5.71° .

    Note that the diagram shows that V2 lags I2 by φ2 or conversely that I2 leads V2. This indicates that the form of I2(t) is

    I2(t) = I2maxsin(Ω2t + φ2) = (0.00697 A)sin(10000t + 5.71°) .

    Since VR2(t) and I2(t) are in phase, the voltage drop over the resistor can now be found using Ohm's Law

    VR2(t) = RI2(t) = (6.97 V) sin(10000t + 5.71°) .

    The net Vout(t) is given by

    Vout(t) = VR1(t) + VR1(t) = (0.995 V) sin(100t + 84.29°) + (6.97 V) sin(10000t + 5.71°) .

    Note that the series RC circuit is a high pass filter. The peak value of high frequency potion of the output is hardly changed from the input value while the low frequency part is ten times smaller.

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  5. A single line carries two signals, as shown in the diagram below, V1 = 10.0sin(100t) and V2 = 7.0sin(10000t) where t is in seconds. An inductor of 1.00 H and a 1000-Ωresistor are connected in series. What is the voltage signal observed over the resistor?

    6. The first step to solving this problem is to realize that V1 and V2 may be treated separately. Let us consider V1 first. Let I1 be the current leaving the power supply.

    The inductor and the resistor are in series so they share the same current, i.e. I1. For an inductor, the current lags the voltage by 90° . The voltage drop over the inductor is related to I1 by VL1 = I1XL1, where XL1 = Ω1L = 100 Ω. The current and voltage drop are in phase for the resistor and VR1 = I1R. The resulting phasor diagram is

    Using the Pythagorean Theorem we find

    Z1 = [(XL1)2 + R2]1/2 = [(100 Ω)2 + (1000 Ω)2]1/2 = 1004.99 Ω.

    From Z1 we can now find the maximum or peak value of I1 using Ohm's Law

    I1max = V1max/Z1 = (10.0 Volts)/(1004.99 Ω) = 0.00995 A.

    The phase angle is

    f 1 = arctan-(XL1/R) = arctan(100/1000) = 5.71° .

    Note that the diagram shows that V1 leads I1 by φ 1 or conversely that I1 lags V1. This indicates that the form of I1(t) is

    I1(t) = I1maxsin(Ω1t - φ 1) = (0.00995 A)sin(100t - 5.71°) .

    Since VR1(t) and I1(t) are in phase, the voltage drop over the resistor can now be found using Ohm's Law

    VR1(t) = RI1(t) = (9.95 V) sin(100t - 5.71°) .

    Next we consider V2(t) and follow the same steps. Let I2 be the current leaving the power supply. Again, the inductor and the resistor are in series so they share the same current, i.e. I2. For an inductor the current lags the voltage by 90° . The voltage drop over the inductor is related to I2 by VL2 = I2XL2, where XL2 = Ω2L = 10,000 Ω. The current and voltage drop are in phase for the resistor and VR2 = I2R. The resulting phasor diagram is

    Using the Pythagorean Theorem we find

    Z2 = [(XL2)2 + R2]1/2 = [(10,000 Ω)2 + (1000 Ω)2]1/2 = 10049.9 Ω.

    From Z2 we can now find the maximum or peak value of I2 using Ohm's Law

    I2max = V2max/Z2 = (7.0 Volts)/(10049.9 Ω) = 0.000697 A.

    The phase angle is

    f2 = arctan(XL2/R) = arctan(10000/1000) = 84.29° .

    Note that the diagram shows that V2 leads I2 by φ2 or conversely that I2 lags V2. This indicates that the form of I2(t) is

    I2(t) = I2maxsin(Ω2t - φ2) = (0.000697 A)sin(10000t - 84.29°) .

    Since VR2(t) and I2(t) are in phase, the voltage drop over the resistor can now be found using Ohm's Law

    VR2(t) = RI2(t) = (0.697 V) sin(10000t - 84.29°) .

    The net Vout(t) is given by

    Vout(t) = VR1(t) + VR1(t) = (9.95 V) sin(100t - 5.71°) + (0.697 V) sin(10000t - 84.29°) .

    Note that the series RL is a low pass filter. The peak value of low frequency potion of the output is hardly changed from the input value while the high frequency part is ten times smaller.

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  6. The AC generator in the diagram below supplies 120 V (RMS) at 60.0 Hz. With the switch open as in the diagram, the current leads the generator emf by 20.0° . With the switch in position 1 the current lags the generator emf by 10.0° . With the switch in position 2 the RMS current is 2.00 A. Find the values of R, L, and C.

    7. The peak value of the voltage is Vmax = (120 Volts)´Ö2. Note that this result only applies to sinusoidal signals. Let w = 2π(60 Hz).

    When the switch is open the inductor, one capacitor, and the resistor are in series. The current, which we shall call I0(t), is common. The impedance of the inductor is XL = w L. The impedance of the capacitor is XC0 = 1/ΩC. The phasor diagram for the voltages is

    Now we are told that I0(t), the generator current, leads the emf by 20° . This means that the circuit is capacitive and that XC0 is larger than XL. The phasor diagram reduces to

    A small amount of vector analysis indicates that

    tan(20°) = [XC0 - XL]/R .;            (1)

    When the switch is in position 1, the two capacitors are in parallel. Hence Ceq = 2C and XC1 = 1/2XC0. As a result of this smaller impedance, the circuit become inductive, with the current I1(t) lagging the emf by 10° . The phasor diagram would look like

    Again, a small amount of vector analysis indicates that

    tan(10°) = [XL - XC1]/R = [XL - 1/2XC0]/R .            (2)

    When the switch is in position 2, the resistor is shorted out and we have only the inductor and one capacitor in series. The phasor diagram is

    which reduces to

    The diagram shows that the emf lags the current by a phase angle of 90° and that Z3 = XC0 - XL. We are told the RMS value of I3 in this case, so I3max = 2Ö2 A.

    Since Vmax = I3maxZ3, our final equation is

    XC0 - XL = Vmax/I3max = 60 .            (3)

    Solving the three equations yields

    R = 60/tan(20°),

    XC0 = 2R[tan(20°) + tan(10°)], and

    XL = R[tan(20°) + 2tan(10°)] .

    The values of the elements are R = 164.8 Ω, C = 14.89 μF, and L =1.957 H.

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  7. An LCR circuit consists of an inductor of 1.25 ´ 10-2 H, a capacitor of 2.75 μF, and a resistor of 5.0 W connected in series with an AC emf. The emf delivers a voltage ε = εmaxsin(Ωt), with εmax = 0.60 V and w = 6.0 ´ 103 rad/s.
    1. What is the impedance of this circuit?
    2. What is the maximum current in this circuit?
    3. What is the phase angle of the current?
    4. What is the natural frequency of this circuit?
    5. What is the ratio of the current in part (b) to the current at resonance?

    The circuit looks like

    1. The impedance of the inductor is XL = ΩL = (6.0 ´ 103 rad/s)(1.25 ´ 10-2 H) = 75 Ω. The impedance of the capacitor is XC = 1/ΩC = 1/(6.0 ´ 103 rad/s)(2.8 μF) = 59.5238 Ω. Since XL, XC, and R are in series, the current is common and the phasor diagram for the voltages looks like

    2. Resolving the vectors will yield the impedance of the circuit

      3. Vector analysis reveals that the impedance is

      .

      To find the current, we use Ohm's Law

      Imax = εmax/Z = (0.60 V)/(16.2638 Ω) = 0.03689 A .

    3. From the last diagram, the phase angle is given by

      4. f = arctan([XL - XC]/R) = 72.0956° .

      Note that the current lags the emf. This makes sense since ZL > ZC which means the circuit is inductive. To be consistent with the given form of ε (t), we should write the current as

      I(t) = (0.03689 A)sin(Ωt - 72.0956°) .

    4. The natural frequency of the circuit is the frequency for which the impedance of the circuit is purely real with no imaginary part. This requires XL = XC. Writing this as ΩL = 1/ΩC, we see that the resonant angular frequency is

      5. w0 = 1/Ö(LC) = 5345 rad/s ,

      and the resonant frequency is

      f0 = Ω0/2π = 851 Hz .

    5. The current at resonance is I0 = εmax/R = (0.60 V)/(5 Ω) = 0.12 A. Therefore the ratio I/I0 = 0.03689/0.12 = 0.307 .

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  8. An LCR circuit consists of an inductor of 1.25 × 10-2 H, a capacitor of 2.75 m F, and a resistor of 5.0 W connected in parallel with an AC emf. The emf delivers a voltage e = emaxsin(wt), with e max = 0.60 V and w = 6.0 × 103 rad/s.

    1. What is the impedance of this circuit?
    2. What is the maximum current produced by the battery?
    3. What is the phase angle of the current?
    4. What is the natural frequency of this circuit?
    5. What is the ratio of the current in part (b) to the current at resonance?

    The circuit looks like

    1. The impedance of the inductor is XL = wL = (6.0 × 103 rad/s)(1.25 × 10-2 H) = 75 W . The impedance of the capacitor is XC = 1/wC = 1/(6.0 × 103 rad/s)(2.8 mF) = 59.5238 W . Since XL, XC, and R are in parallel, the voltage is common and the phasor diagram for the currents looks like

      2. Resolving the vectors will yield the impedance of the circuit

      Vector analysis reveals that the impedance is

      and hence the impedance is Z = 4.9992 W .

    2. To find the current, we use Ohm’s Law

      3. Imax = e max/Z = (0.60 V)/(4.9992 W) = 0.12002 A .

    3. From the last diagram, the phase angle is given by

      4. f = arctan([1/XC - 1/XL] / 1/R) = 0.99303° .

      Note that the current leads the emf. To be consistent with the given form of e (t), we should write the current as

      I(t) = (0.1200 A)sin(wt + .99°) .

    4. The natural frequency of the circuit is the frequency for which the impedance of the circuit is purely real with no imaginary part. This requires 1/XL = 1/XC. Writing this as 1/wL = wC, we see that the resonant angular frequency is

      5. w0 = 1/√(LC) = 5345 rad/s ,

      and the resonant frequency is

      f0 = w0/2p = 851 Hz .

    5. The current at resonance is I0 = emax/R = (0.60 V)/(5 W) = 0.12 A. Therefore the ratio I/I0 = 0.12002/0.12 = 1.0002 .

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  9. For the diagram below, V(t) = 5.00sin(Ωt) Volts, Ω = 1000 s-1, R1 = 100 Ω, R2 = 50.0 Ω, and C = 20.0 μF. Determine the voltage drop over, and current through, each element.

    10. The capacitor has an impedance XC = 1/ΩC = 1/(1000 s-1)(20.0 μF) = 50 Ω. The first step to solving this problem is to get the overall impedance of the circuit.

    We note that R2 and XC are in parallel. Therefore the voltage over each, which we will call V2, is common. The phasor diagram for the currents looks like

    Analyzing the vector diagram yields

    ,

    or ZA = 35.3553 Ω. As well, the phase angle is

    f A = arctan(R2/XC) = 45° ,

    where the current is entering the parallel arms is leading V2. Our circuit diagram has been reduced to R1 and ZA in series. Being in series, elements ZA and R1 carry the same current. The phasor diagram for the voltages is

    Since we are no longer dealing with simple right angles, the analysis of the triangle takes a little more work. The impedance is

    ,

    and the final phase angle is

    .

    Note that the emf lags the current by 11.31° .

    The maximum current is given by Ohm's Law

    Imax = ε max/ZB = (5.00 V)/(127.4755 Ω) = 0.03922 A .

    Thus the current produced by the battery is I(t) = (0.03922 A)sin(Ωt + 11.31°) .

    Now we can work backwards to get V and I for each circuit element. Since R1 and ZA are in series with the battery, they also have current I(t) passing through them. We can find the voltage drop over R1 very simply using Ohm's Law since V and I are in phase for resistors,

    V1(t) = I(t)R1 = (3.922 V)sin(Ωt + 11.31°) .

    Finding the voltage drop over ZA, V2(t), requires more subtlety. First we find V2max using Ohm's Law

    V2max = ImaxZA = (0.3922 A)(35.3553 Ω) = 1.387 Volts .

    Second, we must recall that we discovered that I(t) leads V2(t) by φ A = 45° . That means

    V2(t) = (1.387 V)sin(Ωt + 11.31° - 45°) = (1.387 V)sin(Ωt - 33.69°) .

    Now that we have V2(t), we can find the currents through the parallel arms, I2(t) and IC(t). With R2, V2 and I2 must be in phase, so we just apply Ohm's Law.

    I2(t) = V2(t)/R2 = (1.387 V)sin(Ωt - 33.69°) / 50 Ω= (0.02773 A) sin(Ωt - 33.69°).

    For the capacitor, we apply Ohm's Law to the maximum values, i.e.

    Icmax =V2max/XC = 1.387 V / 50 Ω= 0.02773 A.

    Then we recall that the current "through" a capacitor leads the emf by 90° . Hence

    IC(t) = (0.02773 A)sin(Ωt - 33.69° + 90°) = (0.02773 A)sin(Ωt + 56.31°) .

    In summary,

     

    Voltage (V)

    Current (A)

    Battery

    5sin(Ωt)

    0.03922sin(Ωt + 11.31°)

    R1

    3.922sin(Ωt + 11.31°)

    0.03922sin(Ωt + 11.31°)

    R2

    1.387sin(Ωt - 33.69°)

    0.02773sin(Ωt - 33.69°)

    C

    1.387sin(Ωt - 33.69°)

    0.02773sin(Ωt + 56.31°)

     

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  10. For the diagram below, V(t) = 5.00sin(Ωt) Volts, Ω = 1000 s-1, R = 50.0 Ω, L = 20.0 mH, and C = 20.0 μF. Determine the voltage drop over, and current through, each element.

    11. The capacitor has an impedance XC = 1/ΩC = 1/(1000 s-1)(20.0 μF) = 50 Ω. The inductor has an impedance XL = ΩL = (1000 s-1)(20.0 mH) = 20 Ω. The first step to solving this problem is to get the overall impedance of the circuit.

    We note that R2 and XL are in parallel. Therefore the voltage over each, which we will call V2, is common. The phasor diagram for the currents looks like

    Analyzing the vector diagram yields

    ,

    or ZA = 18.5695 Ω. As well, the phase angle is φA = arctan(R2/XL) = 68.1986° , where the current is entering the parallel arms is lagging behind V2.

    Our circuit diagram has been reduced to XC and ZA in series. Being in series, elements ZA and XC carry the same current. The phasor diagram for the voltages is

    Since we are no longer dealing with simple right angles, the analysis of the triangle takes a little more work. The impedance is

    ,

    and the final phase angle is . Note that the emf lags the current by 78.11° .

    The maximum current is given by Ohm's Law

    Imax = εmax/ZB = (5.00 V)/(33.4767 Ω) = 0.14936 A .

    Thus the current produced by the battery is I(t) = (0.14936 A)sin(Ωt + 78.11°) .

    Now we can work backwards to get V and I for each circuit element. Since XC and ZA are in series with the battery, they also have current I(t) passing through them. Since we have Imax and we know XC, we can find V1max using Ohm's Law

    V1max = ImaxXC = (0.14936 A)(50 Ω) = 7.46788 Volts .

    For capacitors, the current I(t) leads the emf V1(t) by 90° , so the form of V1(t) is

    V1(t) = (7.46788 V)sin(Ωt + 78.11° - 90°) =(7.46788 V)sin(Ωt - 11.89°) .

    The maximum voltage drop over ZA, V2max, is found using Ohm's Law

    V2max = ImaxZA = (0.14936 A)(18.5695 Ω) = 2.7755 Volts .

    Second, we must recall that we discovered that I(t) lags V2(t) by φA = 68.20° . That means

    V2(t) = (2.7755 V)sin(Ωt + 78.11° + 68.20°) = (2.7755 V)sin(Ωt + 146.31°) .

    Now that we have V2(t), we can find the currents through the parallel arms, I2(t) and IL(t). With R, V2 and I2 must be in phase, so we just apply Ohm's Law.

    I2(t) = V2(t)/R = (2.7755 V)sin(Ωt + 146.31°) / 50 Ω= (0.05547 A) sin(Ωt + 146.31°).

    For the inductor, we apply Ohm's Law to the maximum values, i.e.

    ILmax =V2max/XL = 2.7755 V / 20 Ω= 0.13868 A.

    Then we recall that the current through a capacitor lags the emf by 90° . Hence

    IL(t) = (0.13868 A)sin(Ωt + 146.31° - 90°) = (0.13868 A)sin(Ωt + 56.31°) .

    In summary,

     

    Voltage (V)

    Current (A)

    Battery

    5sin(Ωt)

    0.14936sin(Ωt + 78.11°)

    C

    7.4679sin(Ωt - 11.89°)

    0.14936sin(Ωt + 78.11°)

    R

    2.7755sin(Ωt + 146.31°)

    0.05547sin(Ωt + 146.31°)

    L

    2.7755sin(Ωt + 146.31°)

    0.13868sin(Ωt + 56.31°)

     

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  11. For the diagram below, the peak voltage from the power supply is 6.00 Volts. The frequency is f = 300 Hz, R1 = 30.0 Ω, L = 40.0 mH, and C = 25.0 μF. The current from the power supply leads the voltage by 2.50° . Determine R2.

    12. First we compute the reactances. The capacitor has an impedance XC = 1/ΩC = 1/(2π)(300 s-1)(25.0 μF) = 21.22066 Ω. The inductor has an impedance XL = ΩL = 2π(300 s-1)(40.0 mH) = 75.38822 Ω.

    Since R2 is unknown, we can only reduce the circuit to

    The information about the phase must be sufficient to determine R2 when we get to this point.

    The resistor R1 and the capacitor C are in series and thus share the same current, I1. If we call the voltage over the parallel arm VL = V1 + VC, where V1 is the voltage drop over resistor R1 and VC the voltage drop over the capacitor, the phasor diagram looks like

    Analysis of the vector diagram reveals that

    .

    The phase angle is given by

    f A = arctan-1(XC/R1) = arctan-1(21.22066/30) = 35.2739° .

    Note that VL lags I1 by φ A.

    Now the inductor L and ZA are in parallel. They have the same voltage drop. Let's call the current through the inductor IL. The phasor diagram looks like

    Analyzing this vector diagram gives the equivalent impedance

    and the phase angle

    .

    Since φ B is positive, I(t) leads VL(t) by 6.3° .

    The element ZB and R2 are in series and share the same current I(t). The phasor diagram looks like

    Analyzing the vector diagram, we get two equations

    ,

    and

    .

    Since we know ZB and φ B, we can solve the second equation to get our unknown resistance R2 = 67.96 Ω.

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  12. The diagram below shows a circuit with an AC source, V(t) = Vsin(Ωt), and a DC source connected in series with a capacitor and a resistor.
    1. Use the Principle of Linear Superposition to obtain an expression for the current through the resistor as a function of time.
    2. What is the DC offset of the current?
    3. What happens to the DC offset in the limit of large t/RC?
    4. Explain in terms of the RC filter how a suitably chosen capacitor in an AC circuit acts to isolate a load (such as the resistor) from a DC source.

    1. The principle of linear superposition says we can consider the DC source separately from the AC source; the desired current being the sum of the two results.

      2. Consider the DC circuit first.

      The standard result for the DC circuit current is

      IDC(t) = (ε /R)e-t/RC ,

      where RC is the time constant.

      For the AC circuit

      we know that the current IAC(t) is common to C and R. The phasor diagram looks like

      We find the impedance to be Z = {R2 + 1/(ΩC)2}1/2 and the phase angle to be φ = arctan(1/ΩRC). Note that the current leads the voltage by φ . Hence the form of the current is

      IAC(t) = (V/{R2 + 1/(ΩC)2}1/2)sin(Ωt + φ ) .

      Therefore the total current is

      I(t) = IDC(t) + IAC(t) = (ε/R)e-t/RC + (V / R{1 + 1/(ΩRC)2}1/2)sin(Ωt + φ ) .(1)

    2. The DC offset is simply IDC(t).
    3. For large t/RC, IDC(t) ® 0.
    4. Consider equation (1) in the limit of large ΩRC. In this limit {1 + 1/(ΩRC)2}1/2 » {1 + 0}1/2 = 1 and φ = arctan(1/ΩRC) » arctan(0) = 0. Thus the equation becomes

      5. I(t) = (ε /R)e-t/RC + (V/R)sin(Ωt) .

      The exponential part would quickly vanish if RC is also large yielding

      I(t) = (V/R)sin(Ω) .

      Compare this result with the current if there were no capacitor in the circuit. A little thought would indicate that the observed current would be in phase with the AC power supply and have the form

      I(t) = ε /R + (V/R)sin(Ωt) .

      So if we choose C properly, the DC portion of the current is indeed removed.

      We can explain this behaviour in terms of the RC filter. The RC filter is a high pass filter which doesn't effect high frequency terms but does diminish or attenuate low frequency terms. A DC source can be considered as an AC source ε (t) = ε sin(ΩDCt) with ΩDC ® 0, so a DC source is a zero frequency AC source and thus is very strongly diminished.

      The implication of this is that when we solve AC problems that have DC sources and capacitors, the DC sources may be treated as shorts.

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  13. The diagram below shows a circuit with an AC and a DC source connected in series with an inductor and a resistor.
    1. Use the Principle of Linear Superposition to obtain an expression for the current through the resistor as a function of time.
    2. What is the DC offset of the current?
    3. What happens to the DC offset in the limit of large tL/R?
    4. What happens to the pure AC part of the current when ΩL/R >> 1?
    5. Explain in terms of the RL filter how a suitably chosen inductor acts to isolate a load (such as the resistor) from an AC source.

    1. Consider the DC circuit first.

      2. The standard result for the DC circuit current is

      IDC(t) = (ε /R)(1 - e-tL/R),

      where R/L is the time constant.

      For the AC circuit

      we know that the current IAC(t) is common to L and R. The phasor diagram looks like

      We find the impedance to be Z = {R2 + (ΩL)2}1/2 and the phase angle to be φ = arctan(ΩL/R). Note that the current lags the voltage by φ . Hence the form of the current is

      IAC(t) = (V/{R2 + (ΩL)2}1/2)sin(Ωt - φ ) .

      Therefore the total current is

      I(t) = IDC(t) + IAC(t) = (ε/R)(1 - e-tL/R) + (V/ R{1 + (ΩL/R)2}1/2)sin(Ωt - φ ) .(1)

    2. The DC offset is simply IDC(t).
    3. For large tL/R, IDC(t) ® ε /R.
    4. For ΩL/R >> 1, the AC part vanishes, since V/¥ ® 0.
    5. Combining the results of parts (c) and (d), we see that equation (1) reduces to

    I(t) = ε/R ,

    in the limit of large ΩL/R.

    Compare this result with the current if there were no inductor in the circuit. A little thought would indicate that the observed current would be in phase with the AC power supply and have the form

    I(t) = ε/R + (V/R)sin(Ωt) .

    So if we choose L properly, the AC portion of the current is indeed removed.

    We can explain this behaviour in terms of the RL filter. The RL filter is a low pass filter which doesn't effect low frequency terms but does diminish or attenuate high frequency terms. A DC source can be considered as an AC source ε(t) = εsin(ΩDCt) with ΩDC ® 0, so a DC source is a zero frequency AC source and thus is not affected by the filtering action. However L is chosen so that the AC signal is strongly diminished.

    The implication of this is that when we solve problems that have AC and DC sources and inductors, the AC sources may be treated as shorts.

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