Answer: B

 

112. For the circuit in fig. 14 what is the voltage across capacitor when the switch was in position 1 for a long time and closes to position 3 now?

Fig.14

A) 20e-t/RC –5

B) 15e^-t/RC –5

C) 10e^-t/RC –5

D) 20e^-t/RC +5

E) 20e^-t/RC +10

Solution:

V(0)=15 V

V_A =-5 V

V(t)=(V(0)- V_A) e^-t/RC + V_A =20e^-t/RC –5,V

Answer: A

 

113. For the circuit in fig.14 what type of function is applied to RC circuit?

A)pulse

B) step
C)sinusoidal
D) square wave
E)exponent

Answer: B

 

114. For the circuit in fig.14 in what passage the amplitude of capacitor voltage is reached maximum during the transient?

A) 1-3

B) 2-3

C) 3-2

D) 1-2
E) 3-1

Solution:

V(0)=15 V

V_A =-10 V

V(t)=(V(0)- V_A) e^-t/RC + V_A =25e^-t/RC –10,V

For other passages the amplitude is lower than 25V

Answer: D

 

115. For series RL circuit voltage of the source is equal to 10cos10t, V. Current through the circuit is 2cos(10t -30˚), A. Define the value of inductance.

A) 1H

B) 0.75H

C) 0.5H

D) 0.25H

E) 0.1H

Solution:

Z_R=R

Z_L=jwL

Z_eq=R+jwL

Z_eq=V/I=10cos10t/2cos(10t -30˚)=(10∟0˚)/(2∟-30˚)=5∟30˚=4.33+j2.5

Z_L=jwL=j2.5

wL=2.5 w=10rad/s

L=0.25 H

Answer: D

 

 

116. What is angular frequency for the signals v(t)=10cos10t, V?

A) 10Hz

B) 1.59Hz

C) 10rad/s

D)

1.59rad/s

E) 3.18Hz

Solution:

10rad/s, because v(t)= V_A cos(wt+φ) w- angular frequency, then w=10rad/s

Answer: C

 

117. For RC circuit when frequency increases 2 times the impedance of the circuit _______.

A)increases 2 times

B) decreases 2 times

C) increases more than 2 times

D) decreases more than 2 times

E) decreases less than 2 times

Solution:Z₁=ÖR²+(1/wC)², Z₂=ÖR²+(1/2wC)², so if R=2Ω, 1/wC=2Ω, 1/2wC=1Ω, Z₁=Ö2²+2²=Ö8 Ω

Z₂=Ö2²+1²=Ö5 Ω. Z₁/Z₂=Ö8/Ö5=1.26

Answer: E

 

118. Express the signal, whose line spectra are represented in the picture on the right.

A) 8cos(2πf₀t+60˚)+5cos(2π3f₀t-30˚)+ 3cos(2π4f₀t+45˚)

B) 8cos(2πf₀t+60˚)+5cos(2π3f₀t-30˚)

C) 5cos(2πf₀t+60˚)+3cos(2π3f₀t-30˚)+ 10cos(2π4f₀t+45˚)

D) 8cos(2πf₀t+60˚)+5cos(2π3f₀t+30˚)+ 3cos(2π4f₀t+45˚)

E) 8cos(2πf₀t+60˚)+5cos(2π2f₀t-30˚)+ 3cos(2π3f₀t+45˚)

Answer: A

 

119. Find the average power in R=10 Ω if current i(t)=10sin 10t+5sin30t+2sin50t flows through it.

A)320W

B)525W

C)435W

D)480W

E)645W

Solution:

P_av=RIrms²

Irms² =(I₁²+I₂²+I₃²+…+In²)/2= (10²/2+5²/2+2²/2)=64.5

P_av=RIrms² =645,W