Answer: B. 160. Impedances 50 Ω and (-j50) Ω are connected in parallel

160. Impedances 50 Ω and (-j50) Ω are connected in parallel. What is their equivalent impedance?

A).25-j25

B).25+j25

C).70.71∟- 45°

D). 70.71∟45°

E). 35∟45°

Solution:

, Ω

Answer: B

 

161. Define rms value for the function v(t)= 5cos(1000t+30°)

A) 10V B) 5V C) 3.5V D) 7V E) –7V

Solution:

V

Answer: C

162. Nano- means multiplication by

A)10⁶ B). 10^-6 C). 10³ D). 10^-9 E). 10^-12

Answer: D

 

163. We have known R₁, R₂, and R₃ for star. What is R_A for star-to-delta transformation?.

 

 

A)R_A = (R₁ + R₂ + R₃)/ R₁

B). R_A = (R₁ + R₂ + R₃)/R₂ R₁

C) R_A = (R₁R₂ + R₂R₃+R₁R₃)/ R₂

D) R_A = (R₁R₂ + R₂R₃+R₁R₃)/ R₃

E) R_A = (R₁R₂ + R₂R₃+R₁R₃)/ R₁

Solution:

.

Answer: E

 

164. Current i is defined as

A)dw/dt B). dq/dt C). dw/dq D). dq/dv E). dp/dt

Answer: B

 

165. If the zero-input response of parallel RLC circuit is of underdamped form it has got such view:

A)i_L(t)=K₁e^-αtcosβt+K₂e^-αtsinβt B). i_L(t)=K₁e^-αtcosβt+K₂te^-αtsinβt.

C)i_L(t)=K₁e^-α1t+K₂e^-α2t D). i_L(t)=K₁e^-α1t+K₂te^-α2t

E). i_L(t)=K₁e^-αt+K₂te^-αt

Answer: A

 

166. If the zero-input response of parallel RLC circuit is of critically damped form it has got such view:

A)i_L(t)=K₁e^-αtcosβt+K₂e^-αtsinβt B). i_L(t)=K₁e^-αtcosβt+K₂te^-αtsinβt.

C)i_L(t)=K₁e^-α1t+K₂e^-α2t D). i_L(t)=K₁e^-α1t+K₂te^-α2t

E). i_L(t)=K₁e^-αt+K₂te^-αt

Answer: D

 

167. Impedances 50 Ω and (-j50) Ω are connected in series. What is their equivalent impedance?

A).25-j25 B)25+j25 C)70.71∟- 45° D) 70.71∟45° E) 35∟45°

Solution:

70.71∟-45°.

Answer: C

 

168. L=5 mH, ω=10⁶rad/s. Inductive reactance is equal to

A)5kΩ; B)j5kΩ C). – j5kΩ D) j500Ω E). – j500Ω

Solution:

.

Answer: A

 

169. C=5 nF, ω=10⁶rad/s. Capacitive reactance is equal to

A)2kΩ B)j2kΩ C) – j2kΩ D) 200Ω; E) – j200Ω

Solution:

.

Answer: D

 

 

170. C=5 pF, ω=10⁶rad/s. Impedance is equal to

A)200kΩ B)j200kΩ C) – j200kΩ; D) j5000Ω E) – j5000Ω

Solution:

.