Answer: C. 171.For the circuit in fig.20 current of the Norton source is equal to _______ A Fig.20

 

171. For the circuit in fig.20 current of the Norton source is equal to _______ A Fig.20

A)0.555 B)0.1 C)1

D) 0.333 E) 0.666

Solution:

Transform , , , Transform back , , .

Answer: E

 

172. For the circuit in fig.20 Norton equivalent resistance is equal to _______ kΩ.

A)4 B).5 C).7 D).9 E). 12

Solution:

Answer: D

 

173. Define V₀ for the circuit in fig.21. Fig.21

A)V_S *R₄/(R₁+R₄

B) V_S *R₂/(R₁+R₄

C). V_S *R₃/(R₁+R₄)

D) V_S *[(R₂R₃-R₄R₁)/(R₁+R₃)(R₂+R₄)]

E) V_S *R₄/(R₂+R₄)

Answer: D

V₀=V_A-V_B=

(R3*Vs/R1+R3)-(R4*Vs/R2+R4)

 

174. The unit of power is

A)V B). J C) A D) W E) C

Answer: D

 

175. What is correct equation for node A, if we apply node voltage analysis? (G=1/R) (fig.22)

 

 

Fig.22

A)– 0.5Gv_A+3Gv_B-0.5Gv_C=0

B). – 2Gv_A- 0.5Gv_B+3.5Gv_C=0

C)3Gv_A-Gv_B-2Gv_C=i_S

D). - 2.5Gv_A-0.5Gv_B-2Gv_C=i_S

E). 0.5Gv_A+3Gv_B-0.5Gv_C=0

Solution:

.

Answer: C

 

176. Picture A in fig.23 represents input signal of the filter circuit. Define the type of the filter applied if the result of filtration process is shown in picture D as output voltage?

 

Fig.23

A).low-pass filter B).high-pass filter C). band-pass filter

D). stop-band filter E). band-reject filter

Answer: B

 

177. What picture in fig.24 represents band-reject filter circuit?

 

Fig.24

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

Answer: D

178. Define average power consumed by the circuit if applied voltage is v(t)=50+25sin500t, V and the resultant current is i(t)=5+2.23sin(500t-26.6°). Cos26.6°=0.8942

A).250W B)300W C)275 W D)200W E)225W

Solution:

P_av=I₀v₀+ ½I₁v₁ cos(y₁-j₁)+... =>

P_av=250+0.5*25*2.23*0.8942=250+24.9=275 W

Answer: C

 

179. For RC series circuit equivalent impedance is equal to10∟- 30° Ω Define the value of R.

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

=> R = 8.66W