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/2C;

Z_5C= -j/5C

-j/C= 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;

Answer: B

143. What is equivalent capacitance for the circuit below respectively to terminals AB?

A )4mF

B)16mF

C) 8mF

D)6mF

E). 10mF

Solution:

1/ C₁ =1/4+1/4;

C₁=4*4/(4+4)=2mF;

C_eq=2+3+5=10mF;

Answer: E

144. What is equivalent capacitance for the circuit below respectively to terminals AB?

A )1mF

B)2mF

C) 3mF

D)4mF

E)5mF

Solution:

1/ C₁ =1/2+1/2;

C₁=2*2/(2+2)=1mF

C_eq=3+1=4mF;

Answer: D

145. For the circuit on the right define V₀.

A)12V

B)6V

C)4V

D)3V

E)10V

Solution:

V₀=12*R/3R;

V₀=4V;

Answer: C

146. For the circuit in fig.19 the switch was in a neutral position. At t=0 it switches to position 1. What is the current of the transient?

A) 1- e^-500t A B) 0.5(1- e^-500t) A C). 2(1+e^-500t) A

D) 1+e^-500t A E). 0.5(1+e^-500t) A

Solution 1:

100/s =100I(s)+LsI(s)-Li_L(0); Fig.19

i _L(0)=0;

100/s=I(s)(100+0.2s);

I(s)=100/(100+0.2s)s=100/0.2s(500+s)=500/s(500+s);

A/s +B/(500+s) =500/s(500+s);

A500+As+Bs=500;

A500=500; s(A+B)=0;

A=1; A+B=0;

B=-1;

I(s)=1/s - 1/(500+s) ;

i(t)=1-e^-500t A

Solution 2:

i_L(t)=[ i_L(0)- i_L()]*e^-t/Tc+ i_L()

i_L(0)=0

i_L()=Vs/R=100/100=1A

Tc=G_NL=0.2/100=0.002 s

i_L(t)=(0-1)* e^-t/0.002+ 1=1- e^-500t A

Answer: a

147. Main characteristics of sinusoidal waveform are

A)rms value, frequency

B). rms value, frequency, phase angle

C) rms value, frequency, amplitude

D) period, phase angle, rms value

E). amplitude, period, phase angle

Answer: E

148. What is equivalent inductance for the circuit?

A )1 mH

B). 2mH

C). 3mH

D) 4mH

E) 5mH

Solution:

1/ L₁=1/2+1/3+1/6; L₁=1mH;

L_eq=1+4=5mH;

Answer: E

1 49. This is notation of

A) operational amplifier

B) independent current source

C) independent voltage source

D) dependent current source

E) dependent voltage source

Answer: E

150. What is differential equation whose solution is i_L(t)=20e^-1000t-10 mA t≥0.

A)(di_L/dt)+100i_L= -10u(t

B). (di_L/dt)+1000i_L= -10u(t)

C)(di_L/dt)+100i_L= 10u(t)

D). (di_L/dt)+1000i_L= 10u(t)

E) (di_L/dt)+100i_L= 20u(t)

Solution:

i_L(t)=20e^-1000t-10 mA ;

i_L(t)=( I_0- I_A)e^-1000t+ I_A;

I_A = -10mA;

1/ T_C =1000; T_C = 10^-3 s;

T_C(di_L/dt)+i_L= I_A u(t);

10^-3(di_L/dt)+i_L=-10*10^-3u(t);

(di_L/dt)+1000 i_L=-10u(t);

Answer: B

151.W hich nodes are trivial?

A). A, B, E, F

B). E, F, G, H

C). A, B, G, F

D). A, B, E, H

E). C, D, E, F

Answer: B

152. Define peak-to-peak value for the function v(t)= 5cos(1000t+30°)

A) 10V

B) 5V

C) 3.5V D) 7V

E) 2V

Solution:

Answer: A

153. Define equivalent resistance for the circuit below if R₁=R₄=6Ω, R₂=R₅=12Ω, R₃=4Ω, R₆=3Ω.

A )10Ω

B) 12Ω

C).16Ω

D).20Ω

E) 24Ω

Solution:

Ω

Ω

Ω.

Answer: C

154. The reference marks are

A)arrows

B)plus/minus signs

C) arrows and plus/minus signs

D) plus/minus signs or arrows

E) all answers are wrong

Answer: C

155. Main characteristics of exponential waveform are

A) rms value, frequency

B) rms value, frequency, phase angle

C) rms value, frequency, amplitude

D) period, phase angle, rms value

E) amplitude, time constant

Answer: E

156. Current through the inductor for RL circuit is equal to i_L(t)=40e^-200t-20 mA t≥0. What is the initial inductor’s current at t=0?

A)10 mA B). 20 mA C) –10 mA D). –20 mA E). 30 mA

Solution:

mA

Answer: B

157. Conversion of sinusoid v(t)= 20cos(150t-60°) in polar form gives us

A)20∟-150°

B). 20∟150°

C).20∟60°

D). 20∟- 60°

E) 150∟- 60°

Solution:

v(t)= V_Acos(t+ ) => 20∟- 60°.

Answer: D

158. Define amplitude value for the function v(t)= 7cos(1000t+30°)

A).10V

B).5V

C).14V

D).7V

E). –7V

Solution:

v(t)= V_Acos(t+ ) => 7V.

Answer: D

159. Conversion of phasor 7.07- j7.07 at ω=1000 rad/s to sinusoid gives us

A)7.07cos(1000t+45°)

B) 10cos(1000t - 45°)

C). 7.07cos(1000t - 45°)

D) 5cos(1000t+45°)

E). 5cos(1000t+45°)

Solution:

; ; v(t)= 10cos(1000t - 45°).

Answer: B

160. Impedances 50 Ω and (-j50) Ω are connected in parallel. What is their equivalent impedance?

A).25-j25

B).25+j25

C).70.71∟- 45°

D). 70.71∟45°

E). 35∟45°

Solution:

, Ω

Answer: B

161. Define rms value for the function v(t)= 5cos(1000t+30°)

A) 10V B) 5V C) 3.5V D) 7V E) –7V

Solution:

V

Answer: C

162. Nano- means multiplication by

A)10⁶ B). 10^-6 C). 10³ D). 10^-9 E). 10^-12

Answer: D

163. We have known R₁, R₂, and R₃ for star. What is R_A for star-to-delta transformation? .

A)R_A = (R₁ + R₂ + R₃ )/ R₁

B). R_A = (R₁ + R₂ + R₃ )/R₂ R₁

C) R_A = (R₁R₂ + R₂R₃+R₁R₃ )/ R₂

D) R_A = (R₁R₂ + R₂R₃+R₁R₃ )/ R₃

E) R_A = (R₁R₂ + R₂R₃+R₁R₃ )/ R₁

Solution:

.

Answer: E

164. Current i is defined as

A)dw/dt B). dq/dt C). dw/dq D). dq/dv E). dp/dt

Answer: B

165. If the zero-input response of parallel RLC circuit is of underdamped form it has got such view:

A)i_L(t)=K₁e^-αtcosβt+K₂e^-αtsinβt B). i_L(t)=K₁e^-αtcosβt+K₂te^-αtsinβt .

C)i_L(t)=K₁e^-α1t+K₂e^-α2t D). i_L(t)=K₁e^-α1t+K₂te^-α2t

E). i_L(t)=K₁e^-αt+K₂te^-αt

Answer: A

166. If the zero-input response of parallel RLC circuit is of critically damped form it has got such view:

A)i_L(t)=K₁e^-αtcosβt+K₂e^-αtsinβt B). i_L(t)=K₁e^-αtcosβt+K₂te^-αtsinβt .

C)i_L(t)=K₁e^-α1t+K₂e^-α2t D). i_L(t)=K₁e^-α1t+K₂te^-α2t

E). i_L(t)=K₁e^-αt+K₂te^-αt

Answer: D

167. Impedances 50 Ω and (-j50) Ω are connected in series. What is their equivalent impedance?

A).25-j25 B)25+j25 C)70.71∟- 45° D) 70.71∟45° E) 35∟45°

Solution:

70.71∟-45°.

Answer: C

168. L=5 mH, ω=10⁶rad/s. Inductive reactance is equal to

A)5kΩ B)j5kΩ C). – j5kΩ D) j500Ω E). – j500Ω

Solution:

.

Answer: A

169.C=5 nF, ω=10⁶rad/s. Capacitive reactance is equal to

A)2kΩ B)j2kΩ C) – j2kΩ D) 200Ω E) – j200Ω

Solution:

.

Answer: D

170. C=5 pF, ω=10⁶rad/s. Impedance is equal to

A)200kΩ B)j200kΩ C) – j200kΩ D) j5000Ω E) – j5000Ω

Solution:

.

Answer: C

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(₁-₁)+... =>

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 = 8.66

Answer: D

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 = 5

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/(C)

=> C = 1/(*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.F or 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 = 5.2

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 = 3

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 = L

=> L = X_L/) = 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

Answer: D

189. Picture A in fig.27 represents input signal of the filter circuit. Define the type of the filter applied if the result of filtration process is shown in picture C as output voltage?

Fig.27

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

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

Answer: C

190. What i-v characteristic corresponds to the ideal current source? Picture_____

A B C D E

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

Answer: A

191. How many filter prototypes do you know?

A).2 B).3 C).4 D).5 E).6

Answer: C. They are low-pass filter, high-pass filter, band-pass filter, stop-band (band-reject) filter

192. How many equations does it need to prepare if we solve the circuit in fig. 28 applying mesh-current analysis?

Fig.28

A)1 B).2 C).3 D).4 E).5

Answer: A, because two mesh-currents are equal to currents of the sources and are known already.

193. For the circuit below R₁=R₂=100Ω, R₃=50 Ω. What is the value of R_A?

A). 100 Ω B). 200 Ω C). 300 Ω D). 400 Ω E). 250 Ω

Solution:

R_a = (R₁R₂ + R₂R₃ + R₃R₁) / R₁ = (100*100 + 100*50 + 100*50) / 100 = 100 + 50 + 50 = 200 Ω.

Answer: B

194. Three inductors of 2 mH, 3 mH, and 6 mH are connected in parallel. What is equivalent inductance of the circuit?

A).1 mH B)2 mH C).3mH D).4 mH E). 5 mH

Solution:

=> L = 1 mH

Answer: A

195. For picture C in fig.29 C1=50nF, R1=10kΩ, C2=20nF, R2=20kΩ. Define cutoff frequency for the case.

Fig.29

A).100H B)198Hz C).250Hz D).300Hz E).356Hz

Solution:

Hz

Answer: E

196. For low-pass filter R=2.2 kΩ, C=50 nF. What is cutoff frequency for the filter?

A).0.5 kHz B)0.85 kHz C).1.1 kHz D)1.45 kHz E)2.05 kHz

Solution:

Hz

Answer: D

197. For the circuit equivalent impedance is equal to 5∟30° Ω Define the circuit’s admittance.

A).1∟30°S B).1∟-30°S C).0.5∟30°S

D)0.2∟30°S E).0.2∟-30°S

Solution:

Z = 5/2 + (53)j / 2;

Y = 0.2-30° S

Answer: E

198. For series RC circuit voltage of the source is equal to 60cos(200t+20˚), V. What is period for the signal?

A).31.4ms B)100ms C).100μs D).62.8s E). 62.8ms

Solution:

=2f = 2 / T

=> T = 2 /  = 6.28 / 200 = 31.4 ms

Answer: A

199. What picture in fig.30 represents high-pass filter circuit?

Fig. 30

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

Answer: B

200. What picture in fig.30 represents low-pass filter circuit?

Fig. 30

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

Answer: A

ANSWER KEYS.

1	B	51	B	101	D	151	B
2	D	52	C	102	A	152	A
3	E	53	E	103	A	153	C
4	C	54	C	104	E	154	C
5	C	55	E	105	E	155	E
6	C	56	C	106	B	156	B
7	D	57	B	107	D	157	D
8	B	58	D	108	A	158	D
9	D	59	E	109	B	159	B
10	B	60	D	110	A	160	A
11	A	61	A	111	B	161	C
12	C	62	C	112	A	162	D
13	D	63	E	113	B	163	E
14	B	64	C	114	D	164	B
15	C	65	B	115	D	165	A
16	D	66	B	116	C	166	E
17	E	67	B	117	E	167	C
18	A	68	A	118	A	168	A
19	C	69	C	119	E	169	D
20	C	70	B	120	C	170	C
21	B	71	A	121	A	171	E
22	D	72	B	122	C	172	D
23	B	73	E	123	C	173	D
24	B	74	C	124	C	174	D
25	E	75	D	125	E	175	C
26	A	76	D	126	B	176	B
27	B	77	E	127	C	177	D
28	D	78	C	128	C	178	C
29	E	79	C	129	B	179	D
30	C	80	D	130	B	180	A
31	E	81	E	131	A	181	C
32	E	82	A	132	C	182	D
33	C	83	C	133	D	183	E
34	A	84	B	134	A	184	D
35	E	85	E	135	E	185	A
36	C	86	B	136	E	186	C
37	D	87	E	137	C	187	E
38	A	88	A	138	A	188	D
39	B	89	E	139	D	189	C
40	B	90	D	140	A	190	A
41	D	91	C	141	B	191	C
42	B	92	B	142	B	192	A
43	A	93	B	143	E	193	B
44	B	94	E	144	D	194	A
45	D	95	A	145	C	195	E
46	E	96	D	146	A	196	D
47	D	97	E	147	E	197	E
48	A	98	D	148	E	198	A
49	A	99	C	149	E	199	B
50	E	100	A	150	B	200	A