E. A juncture of the terminals of two devices, …a path formed by tracing through a sequence of devices without passing through any node more than once.
Solution:
A node is an electrical juncture of the terminals of two or more devices, but loop is a path formed by tracing through a sequence of devices without passing through any node more than once.
Answer: C.
31. The unit of conductance is
A.C B.A C.J D. Ω E.S
Solution:
Conductance G=1/R, has the unit Siemens, S.
Answer: E.
32. Power, absorbed by resistor, is equal to 400W, when current inside it is 5A. What is the resistance of the resistor? A. 6 Ω B. 10 Ω C. 12 Ω D.20Ω E. 16 Ω
Solution:
P=I2R; R=P/I2=400/25=16Ω
Answer: E.
33. Driving function is __________ produced by ______ because it represents an input that causes a circuit ______.
A. voltage…source…output. B.current …source…output.
C. voltage or current…ideal source…response. D. voltage…ideal source…output
E. current …ideal source…output
Solution:
Driving function is voltage or current produced by ideal source because it represents an input that causes a circuit response.
Answer: C.
34. Power p is defined as
A. dw/dt B. dw/dq C.dq/dt D. dq/dv E. dp/dt
Solution:
Power is defined as dw/dt and is a measure of the rate at which energy is being transferred.
Answer: A.
35. What element(s) can be used as voltage divider?
A. resistor B. diode C. Transistor D. potentiometer E. resistor, potentiometer
Answer: E.
36. Two circuits are said to be equivalent if they have
A. the same view after reduction process
B. identical i-v-characteristics after reduction process
C. identical i-v-characteristics at a specified pair of terminals
D. identical i-v-characteristics
E. identical parameters of elements
Answer: C.
37. The circuit, which is called a Wheatstone bridge, consists of ______ resistors and a source.
A. 2 B.3 C. 4 D.5 E.6
Solution:
Answer: D.
38. When star-connected circuit is balanced and transformed into the delta, the value of side of equivalent delta is equal to ___ of the star’s ray.
A.3 B.1/3 C.2/3 D.1/2 E. 2
Solution:
Let all resistors in star-connected circuit have the value R, then the value of side of equivalent delta is
,
which is equal to 3 times of the star’s ray.
Answer: A.
39. What formula expresses correctly the voltage division rule application for k-th resistor if k of them are connected in series?
A. vk=vtotal*(Req/Rk) B.vk=vtotal*(Rk/Req) C. vk=vtotal*(R1/Req) D. vk=vtotal/(Rk/Req)
E. vk=vtotal*(Rk-1/Req)
Solution:
The Voltage Division
Rule for k-th resistor if k of them are connected in series is
Answer: B.
4
0.
What is the equivalent resistance for the circuit below if all
resistances are equal to R?
(between points A and D)
A.0.5R B.R C.1.5R D.2R E.3R
Solution: 1st way
T
he
values of star sides will be three times less, that is R/3, so after
reducing resistors in series there will be parallel combination of
two resistors with values 4R/3 (R+R/3) in series with one R/3
resistor.
This gives 2R/3+R/3=R.
Answer: B.
2d way
Points B and C are symmetrical respectively to points A and D, so potential of point B equals to potential of point C. Then we can ignore resistor between points B and C. As a result, we have got the circuit, where four resistors of R are connected in two groups of R/2 (two resistors of R in parallel). So resultant value of Req=R/2+R/2=R
41. What formula expresses correctly the current division rule application for k-th resistor if k of them are connected in parallel?
A. ik=itotal*(Req/Rk) B.ik=itotal*(Rk/Req) C. ik=itotal*(R1/Req) D. ik=itotal*(Gk/Geq)
E. ik=itotal*(Geq/Gk)
Solution:
The current division
rule for k-th resistor if k of them are connected in parallel is
Answer: D.
42. 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.
43. Investigate the statements:
I. Node voltage is defined as the voltage between the definite node and the selected reference node.
II. To formulate mesh-current equations we must use branches’ conductances.
III. To apply superposition principle we must define the contribution from each source acting alone.
A. true, false, false B. false, true, false C. false, false, true
D. true, true, false E. true, false, true
Solution:
I. True.
II. False. To formulate mesh-current equations we must use branches’ resistances.
III. False, because the phrase must be continued.
To apply superposition principle we must define the contribution of each source, acting alone, and then take the algebraic sum of the components obtained.
Answer:A.
4
4.
Define Vo for the circuit to the right.
A.1V B. – 1V C. 6V D. 5V E. – 5V
Solution:
Equivalent circuit is
By KVL: 10 + 6 + v0 – 15 = 0; v0 = –1V
Answer: B.
45. What is correct equation for node B (Fig.7), if we apply node voltage analysis? (G=1/R)
A.– 2GvA- 0.5GvB+3.5GvC=0 B. 2GvA- 0.5GvB+3.5GvC=0
C. 0.5GvA+3GvB-0.5GvC =0 D. – 0.5GvA+3GvB-0.5GvC=0
E
.
2.5GvA-0.5GvB -2GvC=iS Fig.
7
Solution:
VB(0.5G+2G+0.5G) –VA(0.5G) –VC(0.5G)=0
–0.5GVA+3GVB–0.5GVC=0
Answer: D.
46. Formulate mesh-current equation for mesh A for the circuit in Fig.8.
A
.
(R3+R4)iB-R3iC=
vS2 B. (R1+R2)iA-R2iC=
vS2
C. (R3+R2)iC-R3iC- R2iA= vS1
D. (R3+R2)iC-R3iC- R2iA= -vS1 E. (R1+R2)iA-R2iC= - vS2
Solution:
iA(R1+R2) – iCR2= –VS2 Fig.8
Answer: E.
47. To apply superposition principle we must ______ in turn all sources, except one. When the current source is _________, we replace it by _________.
A. “turn on”, turned off, short circuit. B. “turn off”, turned on, open circuit.
C. “turn off”, turned on, short circuit. D. “turn off”, turned off, open circuit.
E. “turn off”, turned off, short circuit.
Solution:
To apply superposition principle we must “turn off” all sources in turn. When the current source is turned off, we replace it be open circuit.
Answer: D.
48. Ideal voltage source is described as v=____ and i=____, but ideal current source – i=_____, v=_____.
A. vS, any value, iS, any value B. any value, iS, any value, vS
C. vS, iS, any value, any value D. any value, any value, iS, vS
E. all answers are wrong
Solution:
Ideal voltage source is described as v= vS and i= any value, but ideal current source – i= iS, v= any value.
Answer: A.
49. Find the correct formula for p. p=
A. vi B. dw/dq C.dq/dt D. dq/dv E. di/dt
Solution: p=dw/dt=vi.
Answer: A.
50. The parameters of the Thevenin and Norton equivalent circuits at a given interface can be found as
A. vT=vSC iN=iOC RN=RT=vOC/iSC B. vT=vOC iN=iSC RN=RT=vOC/iOC
C. vT=vSC iN=iOC RN=RT=vSC/iSC
D. vT=vSC iN=iOC RN=RT=vOC/iOC E. vT=vOC iN=iSC RN=RT=vOC/iSC
Solution:
vT=vOC iN=iSC RN=RT=vOC/iSC
Answer: E.
51. Maximum power transfer can be calculated according to the formula
A. Pmax=VT2/4 B. Pmax=iN2RT/4 C. Pmax=VT2/2RT
D. Pmax=VT/4RT E. Pmax=VOC2/2RT
Solution:
Answer: B.
52. Define Thevenin voltage for the circuit in Fig.9.
A.9V B. 8V C. 6V D. 5V E. 3V
Fig. 9
Solution:
E
quivalent
current source will be 15/3*103=5mA
E
quivalent
current source and resistors in parallel will be 5–2=3mA and
6*3/9=2kΩ
Equivalent voltage source and resistors in series will be 3*2=6 and 4+2+3=9kΩ
Load
Answer: C.
53. For the circuit in Fig.9 resistance of the Thevenin equivalent RT is equal to _____
A. 3Ω B. 5kΩ C. 6kΩ D. 8kΩ E. 9kΩ
Answer: E.
54. For ideal open circuit we have got i =_____, R= ____, for ideal short circuit we have got v =_____, R=_____.
A.0, 0, 0, 0 B. ∞, 0, 0, 0 C.0, ∞, 0, 0 D. 0, 0, ∞, 0 E. 0, 0,0, ∞
Solution:
i=0 when R=∞. It’s an open circuit. v=0 when R=0. It’s a short circuit.
Answer: C.
55. To apply superposition principle we must ______ in turn all sources, except one. When the voltage source is _________, we replace it by _________.
A. “turn on”, turned off, short circuit B. “turn off”, turned on, open circuit
C. “turn off”, turned on, short circuit D. “turn off”, turned off, open circuit
E. “turn off”, turned off, short circuit
Solution:
To apply superposition principle we must “turn off” all sources in turn. When the voltage source is turned off, we replace it by short circuit.
Answer: E.
56. In circuit analysis the homogeneity property is called _________, while the additivity property is called_______. Thevenin and Norton equivalent sources are ______ sources.
A. superposition,…. proportionality,… practical
B. superposition,…. proportionality,… ideal
C. proportionality,…superposition,…. practical
D. proportionality,…superposition,…. ideal
E. linearity, … superposition,…. ideal
Solution:
Proportionality: f(kx)=kf(x)
Superposition: f(x1+x2+…+xn) =f(x1)+f(x2)+…+f(xn)
In circuit analysis the homogeneity property is called proportionality, while the additivity property is called superposition. Thevenin and Norton equivalent sources are practical sources.
Answer: C.
57. Proportionality applies when the input and output are ______ . Output_______ is not linearly related to the input _______ .