Alternating-Current Circuits, Solution Manual

Alternating-current circuits Voltages and energies At t = 0 the voltage across the capacitor is V0 cos(0) = V0 . So the voltage across the inductor must be −V0 , because the net voltage change around the loop is zero. The charge on the (top plate of the) capacitor is CV = CV0 cos ωt. The clockwise current is then I(t) = −dQ/dt = ωCV0 sin ωt. This is zero at t = 0, so none of the energy is stored in the LI 2 /2 in the inductor. All of the energy is stored in the CV 2 /2 in the capacitor. This energy equals CV02/2. When ωt = π/2 the voltage across the capacitor is V0 cos(π/2) = 0. So the voltage across the inductor must also be zero. The current at ωt = π/2 is I = ωCV0 sin(π/2) = ωCV0 . Since the voltage across the capacitor is zero, none of the energy is stored in the CV 2 /2 in the capacitor. All of the energy (which we know equals CV02/2) is As a double check, this energy is L(ωCV0 )2 /2 = stored in the LI 2 /2 in the inductor. √ ω 2 LC 2 V02 /2. Using ω = 1/ LC, this becomes CV02/2, as expected. The results are summarized in this table: t=0 t = π/2ω ∆V C V0 0 ∆V L − V0 0 UC CV02 /2 0 UL 0 CV02 /2 Note: At t = 0 you can also work out the voltage across the inductor directly, to double check that it equals −V0 . Using above form of√I(t), the voltage across the inductor is −L(dI/dt) = −ω 2 LCV0 cos ωt. With ω = 1/ LC, this becomes −V0 cos ωt, which equals −V0 at t = 0. However, it’s risky to trust this minus sign. The magnitude is certainly correct, but it’s best to check the sign by thinking about things physically. At t = 0 the current is zero but is increasing in the clockwise direction. The voltage above the inductor must therefore be higher than the voltage below; this difference is what causes the current to increase. Amplitude after Q cycles After Q cycles, the angle ωt equals 2πQ. But from Eq. (8.13) we know that Q = ω/2α. So after Q cycles, the angle ωt equals 2π(ω/2α) = π(ω/α). The time t is therefore given by t = π/α. The exponential factor e−αt that appears in I(t) and V (t) therefore equals e−π as desired. Effect of damping on frequency From Eq. (8.13) we have Q = ωL/R =⇒ R/L = ω/Q, so Eq. (8.9) becomes ω 2 = ω02 − ω2 4Q2 =⇒ ω 2 = ω02 1+ 1 4Q2 . For large Q we can make the approximation, ) )−1/2 ( ( 1 1 1 ω = ω0 1 + ≈ ω0 1 − · 2 . 2 4Q 4Q 2 (528) (529) The fractional shift in ω for Q = 1000 is 1 1 ω0 − ω = = 1.25 · 10−7 , = 8Q2 8 · 106 ω0 (530) or 1.25 · 10−5 percent. The fractional shift in ω for Q = 5 (for which the above Taylor approximation is still quite good) is 1 1 ω0 − ω = = = 0.005, ω0 8 · 25 8Q2 (531) or 0.5 percent, which is still rather small. Decaying signal (a) The impedance of the 105 Ω resistor is much larger than the impedances of the other circuit elements we will find below, so we can neglect the current through the 105 Ω resistor, at least during the initial period right after the switch is closed. During this period we therefore essentially have a series RLC circuit in the right loop. There are two things we can estimate by looking at the given plot: the frequency of oscillation and the rate of decay of the signal. These two things will allow us to calculate C and R, respectively. In cases where the voltage doesn’t immediately become negligible after a few oscillations, the 1/LC term in Eq. (8.9) √ dominates (see Problem 8.5), so the frequency is essentially given by ω = 1/ LC. In the given plot, four cycles are −3 1 completed in 10 sec, so ω = 2π · 4/(10−3 s) = 2.5 · 104 s−1 . The capacitance is then 1 1 C= 2 = = 1.6 · 10−7 F. (532) ω L (2.5 · 104 s−1 )2 (0.01 H) (b) The decay constant is α = R/2L, so the time for the voltage to decrease by a factor of 1/e is t = 2L/R. From the figure, this time is about 0.5 · 10−3 s. Therefore, 2L 2(0.01 H) 2L =⇒ R = = = 40 Ω. (533) t= R 0.5 · 10−3 s t As promised, this is negligible compared with the 105 Ω resistor. Additionally (although at this point we technically haven’t gotten to impedances in this chapter), the impedances of the inductor and capacitor have magnitudes ωL = (2.5 · 104 s−1 )(0.01 H) = 250 Ω and 1/ωC = 1/(2.5 · 104 s−1 )(1.6 · 10−7 F) = 250 Ω. (These are equal because we are using ω 2 = 1/LC.) These are also negligible compared with 105 Ω. But it was fine to just assume that here. 1 In the first printing of the book, the millisec span was drawn a little too long. (c) After a long time, we essentially have just two resistors of 105 Ω and 40 Ω in series with a 20 V dc battery. (No current passes through the capacitor in the eventual direct-current steady state. And there is no voltage drop across the inductor for constant current.) So the voltage across the oscilloscope is ( ) 40 V40 Ω = (20 V) = 0.008 V, (534) 5 10 + 40 or 8 millivolts. The 40 in the denominator is of course inconsequential. Resonant cavity As indicated in Fig. 8.33, let a be the inner radius, b the outer radius, h the height, and s the gap in the capacitor. From Eq. (7.62) with N = 1, the inductance of the single-turn toroid is (µ0 h/2π) ln(b/a). The capacitance of the gap is ϵ0 (πa2 )/s. The resonant frequency is therefore √ √ 2s c 2s 1 = , (535) √ = ω= µ0 ϵ0 ha2 ln(b/a) a h ln(b/a) LC where we have used c2 = 1/µ0 ϵ0 . The fields are shown in Fig. 142. The electric field spans the s gap. From Exercise 6.61, we know that the magnetic field points tangentially around the toroid. The current flows “around” the toroid in the same manner as it did in Exercise 6.61, except that it doesn’t complete a full loop; the charge simply piles up on the end of the inner cylinder and on the corresponding part of the top face of the torus (which completes the capacitor). The charge oscillates back and forth. At the instants when the electric field is maximum, the current is zero so there is no magnetic field. At the instants when the magnetic field is maximum, the charge on the capacitor is zero so there is no electric field. Solving an RLC circuit Let I1 and I2 be the loop currents in the left and right loops, respectively, with clockwise taken to be positive. Let V be the voltage across the capacitor, taken to be positive when the upper plate of the capacitor is positive. The statements involving the three equal voltage drops are V = Q , C V = R′ (I1 − I2 ), V =L dI2 . dt (536) But I1 = −dQ/dt, so if we take the second derivative of the first statement, and also the first derivative of the second statement, we obtain ) ( d2 V dI2 dV 1 dI1 dI2 dI1 − , =− . (537) , V =L = R′ 2 dt dt dt dt dt C dt We can now eliminate I1 and I2 in favor of V by plugging the results for dI1 /dt and dI2 /dt from the first and third equations into the second. The result is ( ) ( ) ( ) V 1 dV 1 dV d2 V d2 V − =⇒ + + V = 0. (538) = R′ −C 2 2 ′ dt dt RC dt LC L dt Using the same trial solution for V as in Eq. (8.4) (and Eq. (8.5)), substituting into Eq. (538), and demanding that the coefficients of sin ωt and cos ωt independently vanish, we obtain, in place of Eq. (8.7), 2αω − ω =0 R′ C and α2 − ω 2 − α 1 + = 0, R′ C LC (539) E B I Figure 142 which give the conditions, α= 1 2R′ C and ω2 = 1 1 − . 4R′2 C 2 LC (540) Alternatively, note that Eq. (538) is identical to Eq. (8.2) if 1/R′ C = R/L, that is, if R = L/R′ C. So we can quickly obtain the results in Eq. (540) by simply replacing the R in Eqs. (8.8) and (8.9) with L/R′ C . Now let’s assume that L, C, and Q are the same in the series and parallel circuits. Since Q = ω/2α, we just need to equate the values of ω/2α for the two circuits. Using Eqs. (8.8) and (8.9), along with Eq. (540), you can show that the values of ω/2α are equal if 1/R′ C = R/L =⇒ R′ = L/RC. In view of the preceding paragraph, this is no surprise, because the time dependence of V is exactly the same in the two circuits if R′ = L/RC. And the energy (which appears in the definition of Q) is proportional to V 2 . Overdamped oscillator From Problem 8.4 the solutions for β are ) ( ) ( √ √ R 4L 1 R R2 4 = 1 ± 1 − . β1,2 = ± − 2 LC 2L R C 2 L L2 (541) √ The roots are real if R ≥ 2 L/C, in which case we have exponentially decaying motion instead of (decaying) oscillatory motion. Note that both roots are positive, as they must be, because a voltage that grows with time would violate conservation of energy. By linearity the most general solution for V is V (t) = Ae−β1 t + Be−β2 t . (542) With R = 600 Ω, L = 10−4 H, and C = 10−8 F, we have 600 Ω R = = 3 · 106 s−1 2L 2 · 10−4 H and 1 4L 4 · 10−4 H = . = 9 (600 Ω)2 (10−8 F) R2 C (543) Therefore, β1 β2 √ ( ) = (3 · 106 s−1 ) 1 + 8/9 = 5.83 · 106 s−1 , √ ( ) 6 −1 = (3 · 10 s ) 1 − 8/9 = 0.172 · 106 s−1 . (544) The constants A and B are determined by the initial conditions. In the setup in Fig. 8.4, the current is zero at t = 0 (because it can’t increase abruptly, due to the inductor), so dQ/dt = 0. And since Q = CV , our initial condition is therefore dV /dt = 0 at t = 0. From Eq. (542), the value of dV /dt at t = 0 is −β1 A − β2 B. This equals zero if B/A = −β1 /β2 ≈ −34. Since β2 is much smaller than β1 , the Be−β2 t term goes to zero much more slowly than the Be−β1 t term. After a microsecond or so, the Be−β2 t term is essentially all that is left. Energy in an RLC circuit The total energy in the capacitor and inductor is CV 2 /2+ LI 2 /2. In the underdamped case, the voltage is given in Eq. (8.10) as V (t) = Ae−αt cos ωt (we can ignore the sin ωt term by adjusting the t (= 0 origin). The current is I = −dQ/dt = −C(dV /dt), so ) we have I(t) = ACe−αt ω sin ωt + α cos ωt . The total energy in the capacitor and inductor is therefore ( ( )2 ) 1 1 1 (545) CV 2 + LI 2 = CA2 e−2αt cos2 ωt + LC ω sin ωt + α cos ωt . 2 2 2 For the overdamped case, the voltage is given in Eq. (8.15) as V (t) = Ae−β1 t +Be−β2 t . The current is then I(t) = −C(dV/dt) = ACβ1 e−β1 t + BCβ2 e−β2 t . The total energy is therefore ( )2 )2 1 1 1 1 ( CV 2 + LI 2 = C Ae−β1 t + Be−β2 t + LC 2 Aβ1 e−β1 t + Bβ2 e−β2 t . (546) 2 2 2 2 For the critically damped √ case, the voltage is given in Eq. (8.16) as V (t) = (A + Bt)e−βt , where β = 1/ LC. This √ form of β follows from Eq. (12.389) in the soluR = 2 L/C. The current is then I(t) = −C(dV/dt) = tion to( Problem 8.4 when ) −βt β(A + Bt) − B . The total energy is therefore Ce ( ( )2 ) 1 1 1 (547) CV 2 + LI 2 = Ce−2βt (A + Bt)2 + LC β(A + Bt) − B . 2 2 2 How fast does the energy decay in these three cases? In the underdamped case, it decays like e−2αt. In the overdamped case, if β2 is the smaller of the two β’s, then for large t the energy decays like e−2β2 t . In the critically damped case, it decays like e−2βt . To show that the energy decays most quickly in the critically damped case, we must show that β > α (so that critical damping decays faster than underdamping), and also that β > β2 (so that critical damping decays faster than overdamping). √ It is indeed true that β > α, because β =√1/ LC and α = R/2L, and the condition for underdamping is precisely R/2L < 1/ LC . Hence α < β . It is also true that β > β2 , because this statement is equivalent to (using the result for β2 in Eq. (12.389)) √ √ R R2 1 1 √ > − ⇐⇒ β > α − α2 − β 2 − 2 2L LC 4L LC √ ⇐⇒ α2 − β 2 > α − β √ √ √ √ ⇐⇒ α−β α+β > α−β α−β √ √ ⇐⇒ α + β > α − β, (548) √ which is indeed true. Note that α > β because R/2L > 1/ LC in the overdamped case, so these square roots are real. For large t, the critically-damped energy in Eq. (547) is essentially equal to Ce−2βt B 2 t2 , where we have used LCβ 2 = 1. Note that the capacitor and inductor have the same energy in the limit of large t. Intuitively, it is believable that the energy decay is quickest for critical damping, because the decay is very slow at both extremes of very light damping and very heavy damping. So the maximum decay rate must occur somewhere in between, and critical damping is as good a guess as any. The decay is slow at the extremes because for fixed L and C, light damping means small R, so there is essentially no resistor in which the energy can dissipate; the energy just sloshes back and forth between the capacitor and inductor, with hardly any overall decrease. And large damping means large R, so there is essentially no current, making the I 2 R term negligible; the energy slowly leaks out of the capacitor. RC circuit with a voltage source This exercise is a special case of the general RLC circuit we solved in Section 8.3. The loop equation here is Q(t) RI(t) + = E0 cos ωt. (549) C Let us replace cos ωt with eiωt , and then guess an exponential solution of the form ˜ iωt satisfies the equation with an eiωt on the right side, then taking Ĩ(t) = Ĩeiωt . If Ie the real part of the entire equation tells us that Re( Ĩeiωt ) satisfies the equation with a cos ωt on the right side. ˜ ˜ = Ĩeiωt , then Q̃(t), which is the integral of I(t), If I(t) equals Ĩeiωt /iω. (There is no need for a constant of integration because we know that Q(t) oscillates around zero.) So we obtain iωt E0 ˜ iωt + Ĩe RIe . (550) = E0 eiωt =⇒ I˜ = R + 1/iωC iωC Getting the i out of the denominator, we can write I˜ in polar form as I˜ = = E0 (R − 1/iωC ) E0 = 2 · (R + i/ωC) R + 1/ω2 C 2 R2 + 1/ω2 C 2 √ E0 E0 · R2 + 1/ω2 C 2 eiϕ = √ eiϕ , R2 + 1/ω2 C 2 R2 + 1/ω2 C 2 (551) where tan ϕ = 1/RωC. The actual current is then ( ) E0 E0 I(t) = Re( Ĩeiωt ) = Re √ cos(ωt + ϕ). eiϕ eiωt = √ 2 R + 1/ω2 C 2 R2 + 1/ω2 C 2 (552) For large ω, the amplitude of the current goes to E0 /R, and the phase ϕ goes to zero. This makes sense, because the capacitor essentially isn’t there (that is, it behaves like a short circuit) because the oscillations happen too quickly for any charge to build up on the capacitor. So we simply have a resistor in series with the voltage source. For small ω, the amplitude of the current goes to zero, and the phase ϕ goes to π/2. In this case, the charge (which has a maximum value of CE0 on the capacitor) sloshes back and forth very slowly, so the current is very small. The resistor essentially isn’t there (the voltage drop IR across it is very small). So we simply have a capacitor in series with the voltage source. And ϕ = π/2 for such a circuit. (The current is ahead of the voltage, because the current reaches its maximum while charge is building up on the capacitor, and then a quarter cycle later the charge reaches its maximum. We are taking Q to be the charge on the top plate of the capacitor, as we did in Section 8.3.) Light bulb The normal current for a 60 watt, 120 volt light bulb is I = P/V = (60 W)/(120 V) = 0.5 A. The resistance of the filament is then R = V /I = (120 V)/(0.5 A) = 240 Ω. (This could also be obtained from P = V 2 /R =⇒ R = V 2 /P .) We want to have the same current, 0.5 A, when the bulb is connected in series with an impedance of iωL, across 240 volts. (We want the same current because the resistor has a fixed resistance, so the power is determined by the current flowing through it; and the power when √ operating normally is 60 watts.) The magnitude of the total impedance is |Z| = R2 + (ωL)2 , so the current will equal 0.5 A if 0.5 = V 240 V V =√ . = √ 2 |Z| R + (ωL)2 (240 Ω)2 + (ωL)2 (553) √ (Ohm’s law works with |Z |; see Eq. (8.77).) Solving for ωL gives ωL = 240 3 Ω = −1 −1 416 Ω. And since ω = 2πν = 2π(60 s ) = 377 s , we have L = (416 Ω)/(377 s−1 ) = 1.10 H. Label the curves The main point here is that in a series RLC circuit, VR , VL , and VC are 90◦ out of phase with each other, while the applied voltage doesn’t (in general) have a nice phase relation with the other three voltages. So we quickly see that the third curve must be the applied E . Also, in a series RLC circuit, VL and VC are 180◦ out of phase. So they must correspond to the 2nd and 4th curves. We can determine which is which by using the fact that VL is 90◦ ahead of VR (which is in phase with the current), which is 90◦ ahead of VC . The 4th curve reaches a maximum 90◦ before the 1st, which in turn is 90◦ ahead of the 2nd. So the 4th is L, the 1st is R, and the 2nd is C. The order of the curves is therefore R, C, E , L. Note that E is slightly ahead of R (or equivalently I). So I is behind E. This means that if E(t) = E0 cos ωt, then the angle ϕ in I(t) = I0 cos(ωt + ϕ) is negative. So from Eq. (8.39), we see that ωL must be larger than 1/ωC. That is, the impedance of the inductor is larger than the impedance of the capacitor. RLC parallel circuit Admittances add in parallel, so 1 i 1 =Y = − + iωC. Z R ωL (554) The given values are R = 103 Ω, C = 5 · 10−10 F, and L = 2 · 10−3 H. For 10 kHz we have ω = 2π(104 s−1 ) = 6.28 · 104 s−1 . Therefore, ( ) 1 i 1 = − + (3.14 · 10−5 )i = 10−3 1 − 7.93i Ω−1 . Z 1000 125.6 (555) Note that the effect of the capacitor is very small. The impedance is Z= 103 (1 + 7.93i) 1 = = (15.7 + 124i) Ω. 10−3 (1 − 7.93i) 1 + 7.932 (556) The inductor dominates the admittance, so Z is roughly equal to iωL. The frequency is low enough so that the inductor lets current through easily compared with the resistor and capacitor. For 10 MHz we have ω = 2π(107 s−1 ) = 6.28 · 107 s−1 . Therefore, ) ( 1 i 1 + (3.14 · 10−2 )i = 10−3 1 + 31.4i Ω−1 . = − Z 1000 1.256 · 105 (557) The effect of the inductor is now negligible. The impedance is Z= 103 (1 − 31.4i) 1 = (1.01 − 31.8i) Ω. = 1 + 31.42 10−3 (1 + 31.4i) (558) The capacitor now dominates the admittance, so Z is roughly equal to 1/iωC. The frequency is high enough so that the capacitor lets current through easily compared with the resistor and inductor. To find the frequency for which |Z| is largest, Eq. (554) gives Z= 1 1/R + i(ωC − 1/ωL) =⇒ |Z| = √ 1 (1/R)2 + (ωC − 1/ωL)2 . (559) This is maximum when the denominator is smallest, that is, when ωC = 1 ωL =⇒ ω = √ 1 1 = √ = 106 s−1 , LC (2 · 10−3 H)(5 · 10−10 F) (560) or equivalently ν = ω/2π = 159 kHz. The maximum impedance is simply |Z|max = R = 1000 Ω. At this frequency, the inductor and capacitor always have equal and opposite currents, so they cancel each other out and effectively don’t exist. Small impedance We have a capacitor in series with the parallel combination of a resistor and an inductor. So the total impedance is Z=− i R(iωL) + . ωC R + iωL (561) √ Setting ω = 1/ LC and combining these two fractions yields Z= L/C √ . R + i L/C (562) If we want this to be small, then we should make R be large. For R → ∞, Z goes to zero. This is because our choice of ω causes the impedances of the inductor and capacitor to exactly cancel. So if R = ∞, the inductor and capacitor are in series, and their impedances add up to zero. The system is on resonance. Any non-infinite value of R will destroy the exact cancelation and produce a nonzero impedance. So we have the counterintuitive result that decreasing the resistance in the circuit increases the impedance. √ In the limit where R ≈ 0 we have Z = −i L/C. In this case, the parallel combination of the R and L has zero impedance, so we are√left with only the impedance of the √ capacitor, which is 1/iωC = −i LC/C = −i L/C, as desired. This case has the largest possible |Z|. Real impedance We have an inductor in series with the parallel combination of a resistor and a capacitor. So the total impedance is Z = iωL + R R(1 − iωCR) 1 = iωL + . = iωL + 1 + ω 2 C 2 R2 1/R + iωC 1 + iωCR (563) Setting the imaginary part of this equal to zero gives ωL(1 + ω 2 C 2 R2 ) − ωCR2 = 0 =⇒ ω 2 = 1 1 − 2 2. R C LC (564) So the answer is “yes,” provided that R2 > L/C, so that ω 2 is a positive quantity. Note that ω = 0 is also a solution that makes the imaginary part of Z be zero. In this case, the capacitor lets through no current (its impedance is infinite), and the inductor is effectively just a short-circuit (its impedance is zero). So we effectively have only the resistor. Equal impedance? Setting the impedances of the two circuits equal to each other gives 1 R(iωL) =R+ iωC R + iωL =⇒ iR2 ωL + Rω2 L2 i =R− . ωC R 2 + ω 2 L2 (565) Equating the real and imaginary parts on the left and right sides of this equation gives Rω2 L2 =R R 2 + ω 2 L2 1 R2 ωL =− . R 2 + ω 2 L2 ωC and (566) The left equation is true if R = 0 or if L = ∞ (or if ω = ∞, but it is understood that we want to find a condition that works for all ω). In the right equation, the negative sign implies that the only way the relation can be true is if both sides are zero. So we must have C = ∞ and either R = 0 or L = ∞. (Again, ω = ∞ works too.) The condition for the two circuits to have equal impedances is therefore: C = ∞ and either R = 0 or L = ∞. In the case where C = ∞ and R = 0, the impedance of both circuits is zero; the capacitor and resistor are short circuits (the impedance of the inductor is nonzero, but that doesn’t matter). In the case where C = ∞ and L = ∞, the impedance of both circuits is R; the capacitor is a short circuit and the impedance of the inductor is infinite. Physically, the reason why there are only a couple special solutions, both of which involve some infinities, is that the magnitude of the impedance of the parallel combination is less than or equal to R, while for the series combination it is greater than or equal to R. Note: if you solve this exercise by multiplying Eq. (565) through by (R + iωL)(iωC ) and simplifying, you must be careful. A spurious solution is introduced, and a valid solution is missed. Zero voltage difference Both branches have the same voltage difference V0 , so the complex currents are given by V0 V0 IA = and IB = . (567) R2 + iωL R1 + 1/iωC The voltage at A is VA = V0 − IA (1/iωC), and at B is it VB = V0 − IB R2 . Therefore, VB − VA = −IB R2 + IA (1/iωC) = = V0 V0 1 · R2 + · R2 + iωL R1 + 1/iωC iωC V0 V0 R 2 + . (568) − iωCR1 + 1 R2 + iωL − This vanishes if R2 (iωCR1 + 1) = R2 + iωL =⇒ iωCR1 R2 = iωL =⇒ R1 R2 = L , C (569) as desired. Given a known capacitance C and known resistors R1 and R2 , we could determine an unknown L by adjusting C, or R1 , or R2 , until we obtain VB − VA = 0. But a real inductor generally has some resistance too, so we effectively have another resistor r in series with R2 and L. This makes things more complicated. L -6 R = 35 Ω C =10 F 15.5V 10.1 V Figure 143 Finding L The setup is shown in Fig. 143. We can quickly determine the amplitude of the current (or the rms value, depending on what the voltmeter is calibrated to read; the final value of L won’t depend on the choice). The frequency is ω = 2π(1000 s−1 ) = 6283 s−1 , so the V0 = I0 |Z| statement for the capacitor alone yields 15.5 V = I0 ωC =⇒ I0 = (15.5 V)(6283 s−1 )(10−6 F) = 0.0974 A. (570) Since the elements are in series, this is the current through all of the components in the circuit. The V0 = I0 |Z| statement for the whole circuit then tells us that (ignoring the units) I0 = V0 |Z| =⇒ 0.0974 = √ 10.1 (35)2 + (ωL − 1/ωC)2 =⇒ ωL − 1/ωC = ±97.6 Ω. (571) Note that there are two roots. We therefore have L= 97.6 1 ± ω2 C ω =⇒ 0.0253 ± 0.0155 =⇒ L = 0.041 H or 0.0098 H. (572) So L could be 41 mH or 9.8 mH. The amplitude of the voltage across the inductor alone is I0 ωL, which gives 25.1 V and 6.0 V for the two possibilities. If we then measure the voltage across the inductor and obtain 25.4 V, the second possibility is ruled out, and we have reasonably good agreement with the computed value of 25.1 V. Equivalent boxes The impedance of the top circuit is (ignoring the units) ( ) 1000 1 + 4 · 10−3 iω + 4000 1 Z = 1000 + = 1 + 4 · 10−3 iω 1/4000 + iω · 10−6 (5000 + 4iω)(1 − 4 · 10−3 iω) 5000 + 4iω = = 1 + 4 · 10−3 iω 1 + 16 · 10−6 ω 2 −3 2 5000 + 16 · 10 ω − 16iω = . 1 + 16 · 10−6 ω 2 (573) The impedance of the bottom circuit is given by 1 Z = = R2 R1 C1 R3 R4 C2 Figure 144 =⇒ Z = 0.64 · 10−6 iω 1 1 1 + = + 6 5000 /(0.64)iω 5000 1250 + 10 8 · 10−4 iω + 1 (1 + 8 · 10−4 iω) + 32 · 10−4 iω 5000(1 + 8 · 10−4 iω) 5000 + 4iω , 1 + 4 · 10−3 iω (574) in agreement with an intermediate step for the top circuit. Now let’s find the general rules for constructing the bottom circuit, given the top circuit. Consider the general circuits shown in Fig. 144. If these two circuits are to have the same impedance for all values of ω , then in particular they must have the same impedance in the ω → 0 and ω → ∞ limits. In the ω → 0 limit, no current goes through the capacitors, so we must have R1 + R2 = R3 (which is indeed true for the given circuits). In the ω → ∞ limit, the capacitor has no impedance, so we must have R1 = R3 R4 /(R3 + R4 ). Using R3 = R1 + R2 and solving for R4 gives R4 = R1 (R1 + R2 )/R2 (which is again true for the given circuits). If the circuits are equivalent at every frequency, they must behave the same way for any pulse or transient. Consider a battery applied across the terminals which charges the capacitor in each circuit. When we remove the battery, the voltage between the terminals will decay with time constant R2 C1 in the top circuit and (R3 + R4 )C2 in the bottom circuit. These times are equal if C2 = C1 R2 /(R3 + R4 ). Using the above values of R3 and R4 , this becomes C2 = C1 R22/(R1 + R2 )2 . This is true for the given circuits. Of course, demanding that the circuits are equivalent in the above three special cases doesn’t actually prove that they are equivalent for all frequencies and scenarios, although it turns out that this is indeed the case. All we’ve shown is that if they are equivalent, then R3 , R4 , and C2 must take on the above values. To be rigorous, we can set the two impedances equal: R1 + 1 1 = . 1 1 1 + + iωC1 R2 R3 R + 1 4 iωC2 (575) If you get everything on one side of the equation and simplify, you will find that there are three different powers of ω. The three coefficients must each be zero, if the equation is to be true for all ω. As you can verify, the term with the smallest power of ω yields the ω → 0 limit above, the term with the highest power yields the ω → ∞ limit, and the middle power (combined with the information from the other two) yields C2 . LC chain The right inductor and Z0 are in series, so they form an impedance of Z0 + iωL, which is then in parallel with the capacitor, all of which is in series with the left inductor. The total impedance of the circuit between the input terminals is therefore 1 Z0 + iωL iωC = iωL + . 1 iZ0 ωC − ω 2 LC + 1 (Z0 + iωL) + iωC (Z0 + iωL) Z = iωL + (576) Setting this equal to Z0 gives ✘ Z0 (iZ0 ωC − ✘ LC + ❆1) ω 2✘ 2 ✘ ✘ Z❩ iZ✘ = iωL(✘ 0 + iωL 0 ωC − ω LC + 1) + ❩ = L(−ω 2 LC + 1) + L √ =⇒ Z0 = (2 − ω 2 LC )(L/C ). (577) √ 2 Z0 is real if ω < 2/LC, as predicted. When ω = 2/LC, we have Z0 = 0. So the circuit looks like the top one shown in Fig. 145, with a short circuit on the right. (You can quickly√double check from scratch that the impedance of this circuit is zero when ω = 2/LC.) This zero impedance makes sense, because the resonant frequency √of each of the √ two halves of the circuit shown on the bottom in Fig. 145 is ω0 = 1/ L(C/2) = 2/LC . Each loop resonates with this frequency, without the need for any applied voltage. Since the impedance is defined by V = IZ, a nonzero current with zero voltage implies zero impedance. L =⇒ Z02C L C L L C/2 C/2 Figure 145 RC circuit (a) The total impedance is Z =R− i i = 2000 Ω − = (2000 − 2650i) Ω. ωC (377 s−1 )(10−6 F) The magnitude is |Z| = √ (578) 20002 + 26502 = 3320 Ω. (b) The rms voltage is V = 120 V, so the rms current is I= 120 V V = = 0.036 A. |Z| 3320 Ω (579) (c) The power dissipated (across just the resistor, of course) is P = I 2 R = (0.036 A)2 (2000 Ω) = 2.6 W. (580) There is no need for a factor of 1/2 since we are using rms values. We can alternatively use the P = V I cos ϕ expression (where these are the rms values). Here V = 120 V, I = 0.036 A, and tan ϕ = 2650/2000 which yields cos ϕ = 0.60. These quantities yield P = 2.6 W, as desired. (d) A voltmeter connected across the resistor will read VR = IR = (0.036 A)(2000 Ω) = 72 V (rms). (581) (rms). (582) A voltmeter connected across the capacitor will read VC = 134 V B A 102 V Figure 146 I = (0.036 A)(2650 Ω) = 95 V ωC (e) The amplitudes of the voltages associated with the above rms values are 102 V and 134 V. The voltages across the resistor and capacitor are 90◦ out of phase, with the resistor ahead of the capacitor. (Remember, in general we have VL ahead of VR ahead of VC .) So the pattern will be an ellipse, as shown in Fig. 146. If the plates are connected in the natural way as shown, then the ellipse is traced out counterclockwise. To see why, consider an instant when the current through the resistor is maximum downward, in which case the right plate of the tube is at a higher potential (so the electrons are deflected that way). The charge on the capacitor is 90◦ out of phase with the current, so there is no charge on the capacitor at this moment. The voltage across the capacitor is therefore zero, so the electrons are at the point A in the figure. A quarter cycle later, the top plate of the capacitor will have maximum charge, in which case the top plate of the tube is at a higher potential (so the electrons are deflected that way). And the current is zero at this moment, so the voltage across the resistor is zero. The electrons are therefore at point B in the figure. We see that the curve on the screen passes point B a quarter cycle after point A. So the curve is traced out counterclockwise. On the other hand, if the connections are made in the reverse manner for either of the elements, then the curve would be traced out clockwise. If both connections are reversed, then the trace reverts back to counterclockwise. Without being told which way the connections are made, there is no way to know the direction of the trace. High-pass filter The current I˜ through the resistor is also essentially the current through the inductor, because the circuit on the right is assumed to have a large impedance. The complex ˜ statement for the voltage between the terminals at A is V0 = I(R ˜ + iωL). Ṽ = IZ ˜ Therefore, And the statement for the terminals at B is Ṽ1 = I(iωL). iωL Ṽ1 = V0 R + iωL  2   ω 2 L2 Ṽ  =⇒  1  = 2 =  V0  R + ω 2 L2 1 . R2 1+ 2 2 ω L (583) This equals 0.1 when R2 /ω2 L2 = 9. At 100 Hz this gives R = 3ω = 3 · 2π · 100 s−1 ≈ 2000 s. L (584) This is satisfied by, for example, R = 100 Ω and L = 0.05 H. The power is proportional to V 2 . If ω ≪ R/L, then we can ignore the “1” in the denominator. So |Ṽ1 /V0 |2 ≈ ω 2 L2 /R2 ∝ ω 2 . Halving ω decreases this by a factor of 4. The physical reason why V1 decreases with decreasing frequency is the following. For very high frequency, the impedance of the inductor is very large. The impedance of the resistor is negligible in comparison, so essentially all of the V0 voltage drop occurs across the inductor, which is what V1 registers. On the other hand, for very low frequency, the impedance of the inductor is very small; it is essentially a short circuit. Therefore, essentially all of the V0 voltage drop occurs across the resistor. Very little occurs across the inductor which, again, is what V1 registers. As in Problem 8.13, adding on another RL loop would square the attenuation effect. The voltage would then be proportional to ω 4 . Parallel RLC power The current I(t) and phase ϕ for the parallel RLC circuit are given in Eq. (8.67). Since tan ϕ can be written as (ωC − 1/ωL)/(1/R), we have cos ϕ = √ 1/R (1/R)2 + (ωC − 1/ωL)2 . (585) Equation (8.84) therefore gives the average power delivered to the circuit as P = = = 1 E0 I0 cos ϕ 2 √ 1 1/R E0 · E0 (1/R)2 + (ωC − 1/ωL)2 · √ 2 (1/R)2 + (ωC − 1/ωL)2 1 E 02 . 2 R (586) The average power dissipated in the resistor is given by Eq. (8.80), where V0 is the voltage across the resistor. But this voltage is simply E0 because we have a parallel circuit. So 1 E02 PR = , (587) 2R in agreement with Eq. (586). Two resistors and a capacitor (a) The impedance of the capacitor is ZC = 1/iωC. But since ω = 1/RC here, we have ZC = −iR. Using the standard rules for adding impedances in series and parallel, the total impedance of the circuit is Z= R(R − iR) 1−i ZR (ZR + ZC ) . = =R 2−i Z R + (Z R + Z C ) R + (R − iR) (588) This can also be written as Z = R(3 − i)/5. (b) The total complex current is Ĩ = √ E0 3 + i E0 10 iϕ E0 2 − i E0 e , = = = R 2 R 2 R 1−i Z where tan ϕ = 1/3. Therefore, √ 10 E0 I0 = 2 R and ϕ = tan−1 (1/3) ≈ 18.4◦ . (589) (590) Formally, ] √ [√ ] [ 10 E0 10 E0 iϕ iωt cos(ωt + ϕ). e e = I(t) = Re Ĩeiωt = Re 2 R 2 R (591) √ (c) Since tan ϕ = 1/3 implies cos ϕ = 3/ 10, the average power dissipated in the circuit is (√ ) 10 E0 1 3E 2 3 1 √ = 0 . E0 I0 cos ϕ = E0 (592) 2 4R 2 2 R 10 Alternatively, we can find the power dissipated by finding the amplitude of the voltage across each of the resistors and then using PR = (1/2)V R2 /R for each. (The resistors are the only places where power is dissipated.) The complex voltage across the right resistor is simply E0 . The complex voltage across the left resistor is E0 ZR /(ZR + ZC ) = E0 /(1 − i), because the complex voltages across the R and C in series are proportional to their impedances. The magnitudes of these two complex√voltages (which are the amplitudes of the two actual voltages) are E0 and E0 / 2. Adding the V R2 /2R powers for each resistor gives 3E02/4R, as above.