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5 Problems with Solutions Introduction to Electrical
Engineering | EE 221
Electrical and Electronics Engineering
West Virginia University (WVU)
3 pag.
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Problem set #9a, EE 221, 10/31/2002 – 10/07/2002
Chapter 10, Problem 1.
A sine wave, f(t), is zero and increasing at t = 2.1 ms, and the succeeding positive maximum of 8.5 occurs at t =
7.5 ms. Express the wave in the form f(t) equals (a) C1 sin (ωt + φ), where φ is positive, as small as possible, and
in degrees; (b) C2 cos (ωt + β), where β has the smallest possible magnitude and is in degrees; (c) C3 cos ωt + C4
sin ωt.
Chapter 10, Solution 1.
(a)
T = 4 (7.5 - 2.1) ms = 21.6 ms, ω = 2π / (21.6ms) = 290.9 rad/s
f(t) = 8.5 sin(290.9t + φ)
at t = 2.1ms ⇒ 0 = 8.5 sin(290.9 * 2.1ms + φ) ⇒ φ = -0.6109 rad + 2π = 5.672 rad = 325°
(b)
8.5 sin (290.0t + 325) = 8.5 cos (290.9t + 325 − 90) = 8.5 cos (290.9t + 235)
= 8.5 cos (290.9t − 125)
(c)
f(t) = 8.5 sin(290.9t + 325) = 8.5 [sin(325) cos(290.9t) + cos(325) sin(290.9t)] =
= 8.5 [-0.574 cos(290.9t) + 0.819 sin(290.9t)] =
= −4.875 cos(290.9t) + 6.963 sin(290.9t)
Chapter 10, Problem 11.
Let vs = 20 cos 500t V in the circuit of Fig. 10.45.
After simplifying the circuit a little, find iL(t).
Hint: Use a Thevenin-equivalent to simplify the
circuit: a voltage source in series with RTH and 20mH. Use
the known relationship (see book p. 312-313 or your notes)
between iL(t) and vs(t) to find the answer.
Chapter 10, Solution 11.
(using 60Ω)
3
vs = 15cos 500t V
4
R th = 5 + 20 60 = 20Ω
At x − x : voc =
∴ Vm = 15V, R = 20Ω, ωL = 10Ω
∴ iL =
10 

cos  500t − tan −1  = 0.6708cos (500t − 26.57) A
20 

202 + 102
15
(Solution for 60kΩ ⇒ Rth = 24.993Ω, Vth = 19.993V, iL = 0.743 cos(500t - 21.807) A
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Chapter 10, Problem 20.
Convert the following to rectangular form:
(a) 7 ∠ -90o
(b) 3 + j + 7 ∠ -17o
o
(c) 14ej15
(d) 1 ∠ 0o
Convert the following to polar form:
(e) –2 (1 + j 9)
(f) 3
Chapter 10, Solution 20.
(a)
7 ∠ -90o = -j 7
(b)
3 + j + 7 ∠ -17o =
= 3 + j + 6.694 – j 2.047 = 9.694 – j 1.047
(c)
14ej15 = 14 ∠ 15o = 14 cos 15o + j 14 sin 15o =
= 13.52 + j 3.623
o
(e) –2 (1 + j 9) = -2 – j 18 = 18.11 ∠ - 96.34o
(f) 3 = 3 ∠ 0o
(d) 1 ∠ 0o = 1
Chapter 10, Problem 27.
In the circuit of Fig 10.52, let the current iL be
expressed as the complex response 20e j (10t +25° ) , and
express the source current is(t) as a complex forcing
function.
Hint: iL(t) → uL(t) and u4Ω(t) → u0.08F(t) → i0.08F(t) → iS(t)
Chapter 10, Solution 27.
iL = 20e j (10t + 25°) A
vL = 0.2
d
[20e j (10t + 25°) ] = j 40e(10t+= 25°) j
dt
vR = 80e j (10t + 25°)
vs = (80 + j 40) e j (10t + 25°) , ic = 0.08(80 + j 40) j10e j (10t + 25°)
∴ ic = (−32 + j 64) e j (10t + 25°) ∴ is = (−12 + j 64) e j (10t + 25°)
∴ is = 65.12e j (10t +125.62°) A
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Chapter 10, Problem 29.
Express each of the following currrents as a phasor:
(a) 12sin (400t + 110°) A
(b) −7 sin 800t − 3cos800t A
(c) 4 cos (200t − 30°) − 5cos (200t + 20°) A.
If ω = 600 rad/s, find the instantaneous value of each of these voltages at t = 5ms
(d) 70∠30° V
(e) - 60 + j 40 V
Chapter 10, Solution 29.
(a)
12 sin (400t + 110°) A → 12∠ 20°A
(b)
−7 sin 800t − 3cos800t → j 7 − 3
= −3 + j 7 = 7.616∠113.20° A
(c)
4 cos (200t − 30°) − 5cos (200t + 20°)
→ 4∠ − 30° − 5∠20° = 3.910∠ − 108.40° A
(d)
ω = 600, t = 5ms : 70∠30° V
→ 70 cos (600 × 5 × 10−3rad + 30°) = −64.95+ V
(e)
ω = 600, t = 5ms :-60 + j 40 V = 72.11∠146.3°
→ 72.11cos (3rad + 146.31°) = 53.75+ V
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