Engineering Electromagnetics
9th Edition
ISBN: 9780078028151
Author: Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher: Mcgraw-hill Education,
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Chapter 10, Problem 10.6P
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7.44 In the circuit of Fig. P7.44, what should the value of L be
104 rad/s so that i(t) is in-phase with u,(t)?
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i(t)
50 Ω
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25 Ω
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Figure P7.44 Circuit for Problem 7.44.
5) An orbiting satellite has both solar panels and a
48-volt battery on board. The instrumentation package
has sent the following data regarding supplied
Coulombs vs. minutes for the 48 volt battery
on the Satellite
Coulombs
3.5
3
25
2
15
05
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0.5 1
1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6
minutes
(a) what is the battery current at t=0-5 minute:
(6) How much power is the battery supplying at 0.5
Minutes?
(c) Between 2 and 3 minutes how much poner does
the battery supply?
(d) Between 3 and 4 minutes what current is produced
by the battery?
7.36 Find the input impedance Z of the circuit in Fig. P7.36 at
0 400 rad/s.
502
3 mH
ww
m
Z→
2 mF
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Figure P7.36 Circuit for Problem 7.36.
9 mH
Chapter 10 Solutions
Engineering Electromagnetics
Ch. 10 - The parameters of a certain transmission line...Ch. 10 - A sinusoidal wave on a transmission line is...Ch. 10 - Prob. 10.3PCh. 10 - A sinusoidal voltage V0, frequency , and phase...Ch. 10 - Two voltage waves of equal amplitude V0 and radian...Ch. 10 - A 50 load is attached to a 50-m section of the...Ch. 10 - Prob. 10.7PCh. 10 - An absolute measure of power is the dBm scale, in...Ch. 10 - A 100-m transmission line is used to propagate a...Ch. 10 - Two lossless transmission lines having different...
Ch. 10 - Two voltage waves of equal amplitude V0, which...Ch. 10 - In a circuit in which a sinusoidal voltage source...Ch. 10 - The skin effect mechanism in transmission lines is...Ch. 10 - A lossless transmission line having characteristic...Ch. 10 - Figure 10.29 See Problem 10.15. For the...Ch. 10 - A 100 lossless transmission line is connected to a...Ch. 10 - Determine the average power absorbed by each...Ch. 10 - The line shown in Figure 10.31 is lossless. Find s...Ch. 10 - A lossless transmission line is 50 cm in length...Ch. 10 - (a) Determine s on the transmission line of Figure...Ch. 10 - Prob. 10.21PCh. 10 - Prob. 10.22PCh. 10 - The normalized load on a lossless transmission...Ch. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - A 75 lossless line is of length 1.2 . It is...Ch. 10 - Prob. 10.27PCh. 10 - The wavelength on a certain lossless line is 10...Ch. 10 - Prob. 10.29PCh. 10 - A two-wire line constructed of lossless wire of...Ch. 10 - In order to compare the relative sharpness of the...Ch. 10 - In Figure 10.17, let ZL=250 and Z0=50. Find the...Ch. 10 - In Figure 10.17, let ZL=100+j150 and Z0=100. Find...Ch. 10 - The lossless line shown in Figure 10.35 is...Ch. 10 - Prob. 10.35PCh. 10 - The two-wire lines shown in Figure 10.36 are all...Ch. 10 - Prob. 10.37PCh. 10 - Repeat Problem 10.37, with, Z0=50 and RL=Rg=25....Ch. 10 - In the transmission line of Figure 10.20, Z0=50,...Ch. 10 - In the charged line of Figure 10.25, the...Ch. 10 - In the transmission line of Figure 10.37, the...Ch. 10 - Figure 10.38 See Problem 10.42. A simple frozen...Ch. 10 - Figure 10.39 See Problem 10.43. In Figure 10.39,...
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- An evening meal is being prepared in a home kitchen containing an electric oven and a microwave oven. The cost for electricity in the home's neighborhood is $0.15 per kilowatt hour. The microwave oven is specified as a 1000 watt unit, while the oven requires 240 volts and uses a current of 30 amperes to cook at 350 degrees Fahrenheit. A frozen meal can be cooked in the microwave oven set on full power in 10 minutes. The same frozen meal cooked in the electric oven set for 350 degrees F takes 40 minutes. (a) How much energy does it take to cook the frozen meal in the microwave at full power and how much does it cost? (b) How much energy does it take to cook the frozen meal in the electric oven at 350 degrees Fahrenheit and how much does it cost?arrow_forwardDon't use ai to answer I will report you answerarrow_forwardAn electrical substation had a sudden discharge arc event lasting 0.005 seconds. The event produced 768,000 volts that conducted 500 amperes to a nearby grounded metal strap and opened a 500 ampere protective breaker. (a) How much power was produced by the electrical discharge? (b) How much energy was in the discharge? (c) How long could a 75 watt light bulb stay lit, if all the energy in the arc was used to operate it?arrow_forward
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- How many atoms are there in a simple cubic unit cell? in a bcc unit cell? in a fcc unit cell? in the unit cell characterizing the diamond lattice?arrow_forwardConsider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go- R 9(1) i(t)↓ After the switches closes, current flows through the circuit and the capacitor begins to discharge. The equation that describes the total voltage in the loop comes from Kirchoff's voltage law: L di(t) + Ri(t)+(0) = 0, (1) where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance, R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of change of the capacitor charge, and dividing by L, we can write the following homogeneous (no input source) differential equation for the charge on the capacitor: 4(1) +29(1)+w79(1)=0, ཀྱི where a= R 2L and The solution to this second order linear differential equation can be written as: 9(1) =Aent - Beat, where (3) (4) (5) A= (81+20)90 +90 (82+20)90 +90 and B= (6)…arrow_forwardConsider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go. R w i(t) q(t) C н After the switches closes, current flows through the circuit and the capacitor begins to discharge. The equation that describes the total voltage in the loop comes from Kirchoff's voltage law: di(t) L + Ri(t) + (t) = 0, dt (1) where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance, R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of change of the capacitor charge, and dividing by L, we can write the following homogeneous (no input source) differential equation for the charge on the capacitor: ä(t)+2ag(t)+wg(t) = 0, (2) where R a 2L and w₁ = C LC The solution to this second order linear differential equation can be written as: where 81= q(t) = Ae³¹- Bel 82 = (3) (4) (5)arrow_forward
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