Answer: E.

64. Conversion of sinusoid v(t)= - 20sin100t in rectangular form gives us

A. 10 – j17.3 B. 20 – j17.3 C. 0 + j20 D. 20 + j17.3 E. 10 – j15.3

Solution:

v(t)= – 20sin100t = – 20cos(100t – 90°)

1^st way: we rotate vector of length 20 by –90 from starting point on 180 degrees on 90° clockwise.

We get vector of length 20 on imaginary axis.

Im

We can describe this vector as 0+j20.

 

2^nd way: V = –20∟–90°

V= –20cos(–90) –j20sin(–90)=0–j20(–1)=0+j20.

Answer: C.

 

65. Conversion of phasor 169∟- 45° at f=60Hz to sinusoid gives us

A. 169cos(150t-45°) B. 169cos(377t-45°) C. 169cos(377t+45°)

D. 169cos(233t-45°) E. 169cos(155t-45°)

Solution:

ω=2πf=2*3.14*60=377

169∟- 45° = 169cos(377t-45°)

Answer: B.

 

66. v₁(t)= 10cos(1000t - 45°) and v₂(t)= 5cos(1000t+30°). Find the sum of two voltages as phasor.

A. V=11.4+j4.57 B. V=11.4 - j4.57 C. V=4.57+j11.4 D. V=4.57-j11.4

E. V=6+j6.4

 

Solution:

10cos(1000t - 45°)=10∟- 45° = 10cos(- 45)+j10sin(-45) = 7.07 – j7.07

5cos(1000t+30°)=5∟30° = 5cos(30)+j5sin(30) = 4.33+j2.5

V=(7.07 – j7.07) + (4.33+j2.5) = 11.4 – j4.57

Answer: B.

 

67. L=12 mH, ω=10⁶rad/s. Impedance is equal to

A. 12kΩ B.j12kΩ; C. - j12kΩ D. j1200Ω E. - j1200Ω

 

Solution:

Z_L=jωL=j12000=j12kΩ

Answer: B.

 

68. What is time constant for the circuit in Fig.10?

A. 3L/5R B. L/3R C. 2L/3R D. 3R/2L E. 2L/5R Fig. 10

Solution:

T_C=G_NL

;

Answer: A.

 

69. Investigate the statements.

I. A phasor is a complex number representing the amplitude and phase angle of a sinusoidal voltage or current.

II. Impedance is the proportionality constant relating phasor voltage and phasor current in linear, two-terminal elements.

III. The sum of phasor currents at a node is zero

A.false, true, true B. true, false, true C. true, true, false D. true, false, false

E. false, false, true

 

Solution:

I. True.

II. True.

III. False. The algebraic sum of phasor currents at a node is zero.