Φ21 Fall 2006 1 AtHW17 SolutionsProblem K33.22 t = 0 s,the currentthe current in the circuit in the gure is 1 2 I0 ?I0 .At what time isSolution: The inductor opposes any change in the current through it.τ = L/R.In an LR circuit, the time constant isThe current dropsexponentially with this time constant.I(t) = e−tR/L −tR/L tR/L t1 I0 2= 1/2 = ln (1/2)Figure 1: An LR circuit for problem K33.22.= ln 2 ( ) = L ln 2/R = 50 × 10−3 H ln 2/ (500 Ω) =2I0 e−t/τ = I0 e−tR/L =6.93 × 10−5 sEnergy within an L-C circuitConsider an L-C circuit with capacitanceC , inductance L, and no voltagesource. As a function of time, the charge on the capacitor is Q(t) and the current through the inductor is I(t). Assume that the circuit has no resistance and that at one time the capacitor was charged. Part A. As a function of time, what is the energy Solution: The energy in an inductor isPart C. What is the total energyinductancestored in the inductor? (In terms ofLandI (t).)stored in the capacitor? (In terms ofCandQ(t).)UL (t) = 12 LI(t)2 .Part B. As a function of time, what is the energy Solution: The energy in a capacitor isUL (t)UC (t) =UtotalUC (t) 21 Q(t) 2 C .stored in the circuit? (In terms of the maximum currentI0andL.)Solution: To solve this, we have to know how the current in an L-C circuit behaves. If there is an initialcurrent owing somehow, it will begin charging the capacitor. The inductor will cause this current to keep owing until the charge builds up enough to oppose the current, at which point the current will stop and the capacitor will be charged. Then, the capacitor will force the current in the opposite direction while the inductor tries to keep the current constant.In the end, the current will ow back and forth sinusoidallythrough the circuit, charging the opposite plates of the capacitor.The mathematical derivation is in thebook, and the current isI(t) = I0 cos (ωt) Without an external voltage source, the frequency will be the resonant frequency of the capacitor and inductor,√ ω = 1/ LC .This tells usUL (t), but we need the charge on the capacitor. Either integrate the Q(t) or use the phasor diagram. The current phasor is to the right. Vc = XC I0 = I0 /ωC and the direction is down. The charge on thecurrent (because it is the derivative of The voltage across the capacitor is capacitor isQ(t) = CVC (t) = CI0 I0 cos(ωt − π/2) = sin (ωt) ωC ω1The total energy is thenUtotal1 I02 1 2 LI0 cos2 (ωt) + sin2 (ωt) 2 2C ω 2 1 2 1 LCI02 LI0 cos2 (ωt) + sin2 (ωt) 2 2 C 1 2 LI 2 0= UL (t) + UC (t) = = =Notice that with no resistance to dissipate the energy, the energy is constant.3A Radio Tuning CircuitA radio can be tuned into a particular station frequency by adjusting the capacitance in an L-C circuit. Suppose that the minimum capacitance of a variable capacitor in a radio is 4.11 pF. Part A. What is the inductanceLof a coil connected to this capacitor if the oscillation frequency of theL-C circuit is 1.58 MHz, corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? Solution: The inductance is found from the resonant frequency formula2ω 2 = (2πf ) =1 LC⇒L=1 2(2πf ) C=1 2(2π1.58 × 106 Hz) (4.11 × 10−12 F )Part B. The frequency at the other end of the broadcast band is 0.533 MHz.capacitanceCmax= 2.47 × 10−3 HWhat is the maximumof the capacitor if the oscillation frequency is adjustable over the range of the broadcastband? Solution: The capacitance is found from the same formula.C=1 2(2πf ) L=1 (2π0.533 ×1062Hz) (2.47 × 10−3 H)2= 3.61 × 10−11 F4Problem K35.36EvaluateVRin the gure at various EMF frequencies.First, let"s think about the phasor diagram.Solution:In the seriescircuit, we know the current is constant, so we draw a current phasor in some direction with lengthI0 .Then we draw thesame direction as the current phasor, with a value of draw theVCphasor at a right angle toVRVR phasor VR = I0 R.in the Then,and in the trailing direction(because voltage lags in a capacitor), which is clockwise from VR . The I length is VC = XC I0 = 0 . Since there is no inductance, we can use the ωC pythagorean theorem to nd the total voltage, which is the vector sumFigure 2: A driven R-C circuit for problem K35.36.of the voltage phasors of the series components.√√ VR2Vtotal =Now that we knowI0 ,we can nd+VC2=( 2(I0 R) +I0 ωC√)2 = I0R2 +1 = E0 ω2 C 2VR . E0 R VR = I0 R = √ R2 + ω21C 2Make sure you understand how this formula came about. This is the imprortant part.
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Now for the numbers. Part A.f = 100 HzSolution:Part B.= 0.994 Vω = 300 · 2π s−1andVR = 2.87 V.ω = 1000 · 2π s−1andVR = 7.07 V.andVR = 9.49 V.f = 3000 HzSolution: Part E.1(200π s−1 )2 (1.59×10−6 F)2f = 1000 HzSolution: Part D.(10 V)(100 Ω) (100 Ω)2 +f = 300 HzSolution: Part C.ω = 2πf = 200π s−1 and VR = √ω = 3000 · 2π s−1f = 10, 000 HzSolution:ω = 10000 · 2π s−1andVR = 9.95 V.Notice that the lower frequencies are blocked by the capacitor while the high frequencies get through to the resistor.3