Answer: C

 

129. Choose the expression, which corresponds to position of the vector (fig.17).

 

 

Fig.17 fig.18

A) A - jB

B) –A + jB

C) A+ jB

D) –A - jB

E) –A ± jB

Answer: B

 

130. Choose the correct sequence of elements (they are connected in series), which corresponds to the vector diagram in fig.18.

A).L, R, L, R

B)C, R, L, R

C). R, L, R, C

D). C, R, L, C

E) R,L,C,L

Answer: B

131. The circuit, which corresponds to vector diagram in fig.18 has ________ character, because current ________ resultant voltage.

A)Resistive-inductive, lags

B) Resistive-capacitive, lags

C) Resistive-capacitive, leads

D) Resistive-inductive, lead

E) Resistive-capacitive, in phase

Answer: A

 

132. An exponential waveform has v(0)=5V and v(2)=1.25V. What are values of v(t) at t=1s and t=4s?

A) 4V, 0.13V

B) 3V, 0.325V

C).2.5V, 0.3125V

D) 1V, 0.285V

E) 1,5V, 0.785V

Solution:

V(0)=5V; V(2)=1.25V; V(1)=?;V(4)=?

v(t)=[v_Ae^-t/Tc_{] U(t)}

v(0)= v_A=5V;

v(2)=5e^{-2/ Tc} =1.25V;

e^{-2/ Tc} =0.25;

T_C=1.44269;

v(1)=5e^-0.693=2.5V; v(4)=5e^-2.773=0.3125V;

Answer: C

 

133. When inductor impedance for the third harmonic equals to j300 Ω its impedance for the sixth harmonic equals to ___ Ω.

A) -j600

B) j1000

C) j1200

D) j600

E) j200

Solution:

Z_3L=3jwL;

Z_6L=6jwL;

jwL = Z_3L/3;

Z_6L=6 Z_3L/3=j600

Answer: D

 

134. One ray of star consists of 2-F capacitor, the second ray of the star is series combination of 2-Ω resistor and 1-F capacitor, and the third ray of the star is the series combination of 1-Ω resistor and 1-H inductance. Define the star’s rays impedances in S-domain.

A) 1+s, 2+1/s, 1/2s

B) 1+2s, 1/s, 2/s

C) 1+s, 2/s,2

D) 1+s, 2/s, 2+1/s

E) 1+2s, 2/s, 1+s

Solution:

First ray: Z₁ = 1/2s;

Second ray: Z₂ = 2+1/s;

Third ray: Z₃= 1+s;

Answer: A

 

 

135. When for RLC series circuit X_L=X_C the circuit works like

A) short circuit

B) open circuit

C) it doesn’t work

D) its impedance is equal to 0

E) its impedance is equal to resistance

Answer: E

 

136. If a waveform has even symmetry the Fourier series consist of _______components

A) sine

B) cosine

C) sine and cosine odd

D) sine and cosine even

E) Dc component and cosine components

Answer: E

 

137. An exponential waveform has v(0)=1.2V and v(3)=0.5V. What are amplitude and time constant for the waveform?

A) 1.2V, 5.13s

B) 3V, 5.13s

C) 1.2V, 3.43s

D) 1.2V, 4.28s

E) 2,5V, 4.28s

Solution:

v(t)=[v_Ae^-t/Tc_{] U(t)}

v(0)= v_A=1.2V

v(3)= 1.2e^{-3/ Tc} =0.5V;

e^{-3/ Tc} =0.42;

-3/ T_C= - 0.875

T_C=3.43s

Answer: C

 

138. What statement is correct?

A). Like charges repel, while unlike charges attract one another.

B) Load is two-port network device

C) Points of an electrical circuit can’t have potentials lower than potential of the ground.

D) Diode id a linear element.

E) The algebraic sum of currents at a node is 0

Answer: A

 

139. When the waveform has some symmetry if dc component is removed the symmetry is called__.

A)odd B)even C)odd or even D)hidden E) half-wave

Answer: D

 

140. When capacitor impedance for the second harmonic equals to –j200 Ω its impedance for the fifth harmonic equals to ___ Ω.

A) -j80

B) j1000

C) -j200

D) -j600

E) j200

Solution:

Z_2C= -j/2wC;

Z_5C= -j/5wC

-j/wC= 2Z_2C;

Z_5C= 2Z_2C/5= -j80;

Answer: A

141. One side of delta consists of 1-F capacitor, the second side of delta is 2- Ω resistor, and the third side of delta is the series combination of 1- Ω resistor and 2-H inductance. Define the delta’s sides impedances in S-domain.

A) 1, 1-2s, 1/s

B) 1+2s, 1/s, 2

C) 1+s, 1/s,2

D) 1+s, 2/s, 2

E) 1+2s, 2/s, 1

Solution:

First side: Z₁ =1/s:

Second side: Z₂ =2

Third side: Z₃ =1+2s

Answer: B

 

142. Voltage of the voltage source is equal to 30 V. Resistance, connected in series with the source, is equal to 1 kΩ. In process of source transformation we defined that current of the equivalent current source is equal to ____ mA

A)20 B) 30 C) 40 D). 50 E) 60

Solution:

I=V/R;

I=30/1000=30mA;