Answer: B.

 

5. 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.

 

6. 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.

 

 

7. Convert the complex number (-12+j16) to polar form

A. 18∟-126.8° B. 20∟126.8° C. 20∟-126.8° D. 22∟-56.8° E. 32∟56.8°

 

Solution:

Vector is situated in the second quadrant.

V =20∟126.8°

Answer: B.

8. Convert the complex number 100∟-120° to rectangular form.

A. 50- j86.6 B. -50-j86.6 C. –50+j86.6 D.33-j75 E. –33-j75

 

Solution:

V_A=100.

v(t)=100cos(ωt - 120°)

V = 100cos(-120°) + j100sin(- 120)= -50-j86.6

Answer: B.

9. Complex number is 20+j11. What quadrant is the phase in?

A. 1 B. 2 C.3 D.4 E.1 or 2

 

Solution:

 

 

Answer: A.

 

10. Convert the complex number 100∟-120° to rectangular form.

A. 50- j86.6 B. -50-j86.6 C. –50+j86.6 D.33-j75 E. –33-j75

 

Solution:

V_A=100.

v(t)=100cos(ωt - 120°)

V = 100cos(-120°) + j100sin(- 120)= -50-j86.6

Answer: B.

 

11. Cyclic frequency for the function v(t)= 5cos(1000t+30°) is equal to

A.1000 Hz B.100Hz C.500Hz D. 318Hz E.159.2Hz

 

Solution:

ω=2πf; f=ω/2π = 1000/2π = 159.2Hz

Answer: E.

12. Cyclic frequency f is equal to

A. 2πT B. 1/2πT C. ω/2π; D. 2πω E. 2π/ω;

 

Solution:

Angular frequency: ω=2πf;

Cyclic frequency: f= ω/2π;

Answer: C.

13. Define V_A for the circuit in Fig.3. V_A=

A.V_S*R4/(R₁+R₄) B. V_S*R₂/(R₁+R₄) C. V_S*R₃/(R₁+R₄) D. V_S*R₃/(R₁+R₃) E. V_S*R₄/(R₁+R₃)

 

Solution:

By Voltage Division Rule: