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Answers to problems in Tutorial 5
- i(0)=0; di(0)/dt=V/L; v(0)=0; dv(0)/dt=0; v(inf)=V; i(inf)=V/R.
- s^2 LC + s L/R + 1 = 0.
- v(t) = V*[1-sqrt(1+sigma^2/omega^2)*exp(sigma*t)*cos(omega*t+phi)], where sigma = -5 Mrad/sec, omega = 3.16 Grad/sec, phi = -0.09 degrees. So finally, v(t) = V*[1-1*exp(-5e6*t)*cos(3.16e9*t - 0.0016 rad)]. Under-damped. For critically damped, R = 15.8 ohms.
- Current in inductor will change instantaneously. L*di/dt will tend to infinity. The very large voltage developed will cause arcing. v(t-T) = V*exp(-t/(RC).
- i(0)=V/R, di(0)/dt=0, v1(0)=V, dv1(0)/dt=0, v2(0)=V, dv2(0)/dt=-V/(R*C2). No arcing. s^3 RLC1C2 + s^2 LC2 + s R(C1+C2) + 1 = 0. i''(0) = -V/(R*C2*L), v1''(0) = 0, v2''(0) = 0.
- vc(0)=0, vc'(0)=200e3 Volts/sec, i(0)=0, i'(0)=1e3 Amps/sec.
- vc(t) = 98.2695*cos(2*pi*60*t - 0.1863 rads).
- s RC + 1 = 0; vc(t) = 98.27*cos(2*pi*60*t - 0.1863 rads) - 96.57*exp(-2000*t).
Answers to problems in Tutorial 6
- 1.2-kVA, 1.09-kW, 507-VAR, pf=0.91 (leading).
- 2.175-kW, 4.575-kW, 2.4-kVA.
- 67.2uF assuming 50-Hz supply; -4.47-kVAR.
- 317.8-W + j*421.7-VAR; 0.0075 Joules; 62.8uF.
- 400-VAR; 88.1uF.
- Current = 50+j*75.9 Amps; reactive power = -15.185-kVAR; 1.21 ohms in
series with 1.7-mF; 4 ohms in parallel with 1.2-mF.
- Current = 50-j*75.9 Amps; reactive power = 15.185-kVAR; 1.21 ohms in
series with 5.8-mH; 4 ohms in paprallel with 8.4-mH.
Answers to problems in Tutorial 7
- Center frequency at 1.5915-GHz, range of frequencies from 1.304-GHz to 1.942-GHz.
- Center frequency at 1.66-GHz, range of frequencies from 1.235-GHz to 2.19-GHz.
Answers to problems in Tutorial 8
- Assume 440-V is line voltage. 38.1 Amps through each resistor. Line current: 66 Amps.
- Most likely impedance: 40 angle -20 degrees. 6*sqrt(3) angles 110 degrees, -10 degrees and -130 degrees.
- 328:10; 346-mAmps.
- 480*sqrt(2) angle 120 degrees; 10*sqrt(2)/sqrt(3) angles 0, 120, -120 degrees; phase impedance = 48*sqrt(6) ohms; 11.75kW always.
- Voltage at Bhakra = 140.95kV; power lost = 131.3kW; 0.95%
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