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http://iml.umkc.edu/physics/wrobel/phy250/homework.htm Homework 8 chapter 33: 23, 33, 43 Problem 33.23 A person is working mear the secondary coil of a transformer as shown in Figure 33.25. The primary voltage is 120 V at 60 Hz. The capacitance Cs, which is the stray capacitance between the hand and the secondary winding, is 20 pF. Assuming the person has a body resistance to ground Rb = 50 kΩ, determine the rms voltage across the body. 5,000 V The secondary circuit consists of a “capacitor” with capacitance Cs=20μF, a “resistor” with resistance Rb=50kΩ and a source of electromotive force (the secondary coil of the transformer). The source produces sinusoidal potential difference with rms value of 5000 V and frequency of 60 Hz. At this frequency the impedance of the “capacitor” is ZC = 1 2πf ⋅ Cs and the impedance of the “resistor” is equal to its resistance ZR = R b . Since the “elements” are connected in series, the equivalent resistance of the system (of the two elements) is Z = R 2b + 1 2πf ⋅ Cs 2 The rms current in the “resistor” (and the circuit) is therefore I rms = Vrms = Z Vrms 1 R 2b + 2πf ⋅ Cs . From Ohm’s law, the rms voltage across the body (resistor) is Vb, rms = I b, rms R b = R b = Vrms ⎛ 1 ⎞ ⎟⎟ R 2b + ⎜⎜ ⎝ 2πf ⋅ Cs ⎠ Vrms ⎛ ⎞ 1 ⎟⎟ 1 + ⎜⎜ 2 f C R π ⋅ ⋅ ⎝ s b ⎠ 2 = = 1.88V 3 2 = 5kV ⎞ ⎛ 1 1 + ⎜⎜ ⎟⎟ 2 60 Hz 20 pF 50 k π ⋅ ⋅ ⋅ Ω ⎠ ⎝ 2 Problem 33.33 diode A diode is a device that allows current to pass in only one direction (indicated by the arrowhead). Find, in terms of Vrms and R, the average power dissipated in the diode circuit shown in Figure P33.35. 2R R R R diode V When the left side of the diode generator has a higher 2R potential, the top diode R conducts (forward bias), and the bottom diode does not R (reverse bias). The remaining circuit of three resistors acts like a single resistor. The + resistors R are connected in V series and the pair is connected in parallel to the resistor 2R. Therefore the equivalent resistance of the system is 1) R +eq 1 1 ⎞ = ⎛⎜ + ⎟ ⎝ 2R R + R ⎠ −1 =R Knowing the rms value of the voltage we could find the power in this circuit if the diode was not present. However, the presence of the diode reduces the power to half its value 4 (because only half of the time current would flow in the circuit). Therefore 2 1 Vrms + 2) P = ⋅ 2 R 2R When the polarity of the emf is R reversed, the top diode is in R R reverse bias and the bottom diode diode is in forward bias. This time the equivalent resistance of + the system of resistors is V different 3) −1 1 1 ⎞ 7 ⎛ − R = R+⎜ + ⎟ = R ⎝ R R + 2R ⎠ 4 With the same arguments as before, the power dissipated at this polarity is 2 1 Vrms − 4) P = ⋅ . R 2 175 The overall (time) average power is 2 2 Vrms ⎛ 4⎞ 11 Vrms + − P= P +P = ⋅ ⎜1 + ⎟ = ⋅ 2R ⎝ 7⎠ 14 R 5 Problem 33.43 A transmission line that has a resistance per unit length of 4.5⋅10-4 Ω/m is to be used to transmit 5 MW over 400 miles (6.44⋅105 m). The output voltage of the generator is 4.5 kV. (a) What is the line loss if a transformer is used to step up the voltage to 500 kV? (b) What fraction of the input power is lost to the line under these circumstances? (c) What difficulties would be encountered in attempting to transmit the 5 MW at the generator voltage of 4.5 kV 500 kV or 4.5 kV λ = 4.5⋅10-4 Ω/m 5 MW L = 6.44⋅105 m a) In order to send out power of P = 5MW at 500 kV potential difference, the current in the grid (transmission lines) must be I rms, a = P Vrms, a = 10A The resistance of the two wires of the transmission lines (connected is series) is R = 2 ⋅ L ⋅ λ = 580Ω Hence the loss of power in the lines (dissipated in the lines) is Pa = I 2rms ⋅ R = (10A ) 580Ω = 58kW 2 b) Relative small fraction is lost in the lines Pa 58kW = = 0.01 = 1% P 5mW 6 c) If the power were send at 4.5 kV potential difference, the current in the grid (transmission lines) should be much larger I rms, c = P Vrms, c = 5MW = 1.1kA 4.5kV But even if nothing were connected at the end of the lines, with the potential difference of 4.5 kV, the current in the lines would be I rms, max = Vrms, c 4.5kV = = 77A 58Ω R It would be impossible to send 5 MW of power into the lines. 7