Answer: E
150. What is differential equation whose solution is iL(t)=20e-1000t-10 mA t≥0. A)(diL/dt)+100iL= -10u(t B). (diL/dt)+1000iL= -10u(t) C)(diL/dt)+100iL= 10u(t) D). (diL/dt)+1000iL= 10u(t) E) (diL/dt)+100iL= 20u(t) Solution: iL(t)=20e-1000t-10 mA; iL(t)=(I0- IA)e-1000t+ IA; IA = -10mA; 1/ TC =1000; TC = 10-3 s; TC(diL/dt)+iL= IA u(t); 10-3(diL/dt)+iL=-10*10-3u(t); (diL/dt)+1000 iL=-10u(t); Answer: B
151. A). A, B, E, F B). E, F, G, H C). A, B, G, F D). A, B, E, H E). C, D, E, F Answer: B
152. Define peak-to-peak value for the function v(t)= 5cos(1000t+30°) A) 10V B) 5V C) 3.5V E) 2V Solution: Answer: A
153. Define equivalent resistance for the circuit below if R1=R4=6Ω, R2=R5=12Ω, R3=4Ω, R6=3Ω.
B) 12Ω C).16Ω; D).20Ω E) 24Ω Solution:
Answer: C 154. The reference marks are A)arrows B)plus/minus signs C) arrows and plus/minus signs D) plus/minus signs or arrows E) all answers are wrong Answer: C
155. Main characteristics of exponential waveform are A) rms value, frequency B) rms value, frequency, phase angle C) rms value, frequency, amplitude D) period, phase angle, rms value E) amplitude, time constant Answer: E 156. Current through the inductor for RL circuit is equal to iL(t)=40e-200t-20 mA t≥0. What is the initial inductor’s current at t=0? A)10 mA B). 20 mA C) –10 mA D). –20 mA E). 30 mA Solution:
Answer: B
157. Conversion of sinusoid v(t)= 20cos(150t-60°) in polar form gives us A)20∟-150° B). 20∟150° C).20∟60° D). 20∟- 60° E) 150∟- 60° Solution: v(t)= VAcos(wt+ Answer: D
158. Define amplitude value for the function v(t)= 7cos(1000t+30°) A).10V B).5V C).14V D).7V E). –7V Solution: v(t)= VAcos(wt+ Answer: D
159. Conversion of phasor 7.07- j7.07 at ω=1000 rad/s to sinusoid gives us A)7.07cos(1000t+45°) B) 10cos(1000t - 45°) C). 7.07cos(1000t - 45°) D) 5cos(1000t+45°) E). 5cos(1000t+45°) Solution:
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