homework 9 [solutions] - kyle montgomery, phd - sq14 - home… · homework 9 [solutions] due by...

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1 / 14 ENG17, Sec. 2 (Montgomery) Spring 2014 Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored in the circuit below at the time the switch is opened. a. Derive the differential equation that governs the behavior of i 2 is L 1 = 4 H, L 2 = 16 H, M = 2 H, and R o = 32 Ω. b. Show that when i g = 8 8e -t A, t ≥ 0, the differential equation derived in (a) is satisfied when i 2 = e -t e -2t A, t ≥ 0. c. Find the expression for the voltage v 1 across the current source. d. What is the initial value of v 1 ? Does this make sense in terms of known circuit behavior?

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Page 1: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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ENG17, Sec. 2 (Montgomery)

Spring 2014

Homework 9 [Solutions]

due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131)

1. (P6.39) There is no energy stored in the circuit below at the time the switch is opened.

a. Derive the differential equation that governs the behavior of i2 is L1 = 4 H,

L2 = 16 H, M = 2 H, and Ro = 32 Ω.

b. Show that when ig = 8 – 8e-t A, t ≥ 0, the differential equation derived in (a) is

satisfied when i2 = e-t – e

-2t A, t ≥ 0.

c. Find the expression for the voltage v1 across the current source.

d. What is the initial value of v1? Does this make sense in terms of known circuit

behavior?

Page 2: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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2. (P6.44) The self-inductances of the coils below are L1 = 18 mH and L2 = 32 mH. If the

coefficient of couple is 0.85, calculate the energy stored in the system in mJ when

a. i1 = 6 A, i2 = 9 A;

b. i1 = -6 A, i2 = -9 A;

c. i1 = -6 A, i2 = 9 A; and

d. i1 = 6 A, i2 = -9 A;

Page 3: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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3. (P6.46) Two magnetically coupled coils have self-inductances of 60 mH and 9.6 mH,

respectively. The mutual inductance between the coils is 22.8 mH.

a. What is the coefficient of coupling?

b. For these two coils, what is the largest value that M can have?

c. Assume that the physical structure of these coupled coils is such that P1 = P2.

What is the turns ratio N1 / N2 if N1 is the number of turns on the 60 mH coil?

Page 4: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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4. (P9.76) The sinusoidal voltage source in the circuit below is operating at a frequency of

200 krad/s. The coefficient of coupling is adjusted until the peak amplitude of i1 is

maximum.

a. What is the value of k?

b. What is the peak amplitude of i1 if vg = 560 cos (2 x 105t) V?

Page 5: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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Page 6: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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5. (P9.82)

a. Show that the impedance seen looking into the terminals a,b in the circuit below

is given by the expression

(

)

b. Show that if the polarity terminal of either one of the coils is reversed that

(

)

Page 7: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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6. (P10.6) Find the average power dissipated in the 30 Ω resistor in the circuit below if

ig = 6 cos 20,000t A.

Page 8: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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7. (P10.9) The circuit below has the following sinusoidal source voltages

a. Calculate the real and reactive power associated with each circuit element.

b. Verify that the average power generated equals the average power absorbed.

c. Verify that the magnetizing vars generating equal the magnetizing vars absorbed.

Page 9: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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8. (P10.13) Find the rms value of the periodic current shown below.

Page 10: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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9. (P10.17) Find the average power, the reactive power, and the apparent power absorbed by

the load in the circuit below is ig equals 40 cos 5000t mA.

Page 11: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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10. OPTIONAL: (P10.25) The three parallel loads in the circuit below can be described as

follows: Load 1 is absorbing an average power of 7.5 kW and 9 kVAR of magnetizing

vars; load 2 is absorbing an average power of 2.1 kW and generating 1.8 kVAR of

magnetizing reactive power; load 3 consists of a 48 Ω resistor in parallel with an

inductive reactance of 19.2 Ω. Find the rms magnitude and the phase angle of Vg if Vo =

480 0° V (rms).

Page 12: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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11. OPTIONAL: (P10.37)

a. Find the average power dissipated in each resistor in the circuit below.

b. Check your answer by showing that the total power developed equals the total

power absorbed.

Page 13: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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12. OPTIONAL: (P10.42) The 9 Ω resistor in the circuit below is replaced with a variable

impedance Zo. Assume Zo is adjusted for maximum average power transfer to Zo.

a. What is the maximum average power that can be delivered to Zo?

b. What is the average power developed by the ideal voltage source when maximum

average power is delivered to Zo?

Page 14: Homework 9 [Solutions] - Kyle Montgomery, PhD - SQ14 - Home… · Homework 9 [Solutions] due by 11:45am, Tuesday 6/3/14 (in HW box in Kemper 2131) 1. (P6.39) There is no energy stored

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13. OPTIONAL: (P10.56) The impedance ZL in the circuit below is adjusted for maximum

average power transfer to ZL. The internal impedance of the sinusoidal voltage source is

4 + j7 Ω.

a. What is the maximum average power delivered to ZL?

b. What percentage of the average power delivered to the linear transformer is

delivered to ZL?