Answer: D. 180.For RC series circuit equivalent impedance is equal to10∟- 30° Ω Define the value of capacitive reactance.

 

180. For RC series circuit equivalent impedance is equal to10∟- 30° Ω Define the value of capacitive reactance.

A).5 Ω; B) 6.5 Ω C).7 Ω D)8.66 Ω E).9.5 Ω

Solution:

Z=R – jXc=10∟- 30°

Z=8.66-5j

=> Xc = 5W

Answer: A

 

181. For RC series circuit equivalent impedance is equal to10∟- 30° Ω Define the value of C if angular frequency. ω=10³rad/s.

A).0.5 mF B).0.6 mF C)0.2 mF D).0.8 mF E).0.3 mF

Solution:

Z=R – jXc=10∟- 30°

Z=8.66-5j

Xc =1/(wC)

=> C = 1/(w*Xc) = 1/(1000*5) = 0.2*0.0001 F = 0.2 mF

Answer C

 

182. For picture C in fig.25 C1=50nF, R1=10k Ω, C2=20nF, R2=.20k Ω. Define passband for the case.

 

Fig.25

A)20-100Hz B).118-198Hz C).250-350Hz

D)318-398Hz E)400-480Hz

Solution:

Hz

Hz

Answer: D

183. What i-v characteristic corresponds to the ideal voltage source? Picture_____

A B C D E

 

A).A B).B C)C D)D E)E

Answer: E

 

184. For the circuit in fig.26 define number of resistors where current flows when the switch is ON and OFF

 

 

A) 5,5

B) 5,4

C) 4,4 Fig.26

D) 4,5

E) 5,3

Answer: D

 

185. For RL series circuit equivalent impedance is equal to 6∟30° Ω Define the value of R.

A).5.2 Ω;B).4.3 Ω C).3 Ω D).2.6 Ω E).2.3 Ω

Solution:

Z = R +jX_L =6∟30°

Z = 6*cos30 + (6*sin30)*j = 5.2 + 3j

=> R = 5.2W

Answer: A

 

186. For RL series circuit equivalent impedance is equal to 6∟30° Ω Define the value of inductive reactance.

A).5.2 Ω B).4.3 Ω C).3 Ω; D)2.6 Ω E)2.3 Ω

Solution:

Z = R +jX_L =6∟30°

Z = 6*cos30 + (6*sin30)*j = 5.2 + 3j

=> X_L = 3W

Answer: C

 

187. For RL series circuit equivalent impedance is equal to 6∟30° Ω Define the value of L if angular frequency. ω=10³rad/s.

A).5 mH B).6 mH C).2 mH D).8 mH E)3 mH

Solution:

Z = R +jX_L =6∟30°

Z = 6*cos30 + (6*sin30)*j = 5.2 + 3j

X_L = wL

=> L = X_L/w) = 3j/(j*1000) = 3 mH

Answer: E

 

188. Define average power consumed by resistor of 5 Ω if applied voltage is v(t)=100+22.4sin500t+ 4.11sin1500t.

A).2500W B)3000W C).2750 W D).2052W E)2257W

Solution:

P_av = V_rms² / R = (V₀² + 0.5V₁²+…) / R =

= (10000 + 0.5*22.4² + 0.5*4.11²) / 5 = (10000 + 250.88 + 8.44605) / 5 = 2051.8 => 2052W