Answer: D. 87. When delta-connected circuit is balanced and transformed into the star, the value of ray of equivalent star is equal to ___ of the delta’s side.

 

87. When delta-connected circuit is balanced and transformed into the star, the value of ray of equivalent star is equal to ___ of the delta’s side.

A.3 B.1/3 C.2/3 D.1/2 E. 2

 

Solution:

Let all resistors in delta-connected circuit have the value R, then the value of ray of equivalent star is

, which is equal to 1/3 of the delta’s side.

Answer: B.

 

88. What is correct equation for node B (Fig.7), if we apply node voltage analysis? (G=1/R)

A.– 2Gv_A- 0.5Gv_B+3.5Gv_C=0 B. 2Gv_A- 0.5Gv_B+3.5Gv_C=0

C. 0.5Gv_A+3Gv_B-0.5Gv_C =0 D. – 0.5Gv_A+3Gv_B-0.5Gv_C=0

E. 2.5Gv_A-0.5Gv_B -2Gv_C=i_S Fig. 7

 

Solution:

V_B(0.5G+2G+0.5G) –V_A(0.5G) –V_C(0.5G)=0

–0.5GV_A+3GV_B–0.5GV_C=0

Answer: D.

 

 

89. What statement is wrong?

A. Maximum current available at the interface is the source short-circuit current.

B. When R_L=R_T exists, the source and load are said to be matched.

C. Voltage division and current division are examples of proportionality.

D. Thevenin resistance is equal to input resistance of the circuit.

E. Thevenin and Norton equivalent sources are practical sources.

 

Solution:

D – wrong., because Thevenin resistance is equal to lookback resistance of the circuit.

Answer: D.

 

90. _________ can be defined if we consider the circuit respectively to the interface when all sources are turned off.

A. Thevenin resistance B. input resistance C. lookback resistance

D. Norton resistance E. answers A, C, D are correct

 

Solution:

Thevenin, Norton and lookback resistance can be defined if we consider the circuit respectively to the interface when all sources are turned off.

Answer: E.

 

 

91. v₁(t)= 10cos(1000t - 45°) and v₂(t)= 5cos(1000t+30°). Find the sum of two voltages as phasor.

A. V=11.4+j4.57 B. V=11.4 - j4.57 C. V=4.57+j11.4 D. V=4.57-j11.4

E. V=6+j6.4

 

Solution:

10cos(1000t - 45°)=10∟- 45° = 10cos(- 45)+j10sin(-45) = 7.07 – j7.07

5cos(1000t+30°)=5∟30° = 5cos(30)+j5sin(30) = 4.33+j2.5

V=(7.07 – j7.07) + (4.33+j2.5) = 11.4 – j4.57