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Unformatted text preview: ’ ‘ Agricultural and Biosystems Engineering Dept. 9/21/06
Iowa State University SKW Namr: Egg) TSM 363  Exam 1 Part 1  Formula Sheet (one side 8.5”x11”)
Multiple Choice & Short Essay: 2 points each (40%) 1) In an AC circuit, power is used by
@ resistances
b) capacitances
c) inductances
d) all of the above 6) (a) & (b)
D (a) & (c)
g) (b) & (c) 2) When an extra series load is added to a series circuit,
a) voltages across and currents through other loads are not changed.
b) voltages change, but currents remain the same.
A c currents change, but voltages remain the same.
é voltages across and currents through other loads change. 3) Resistance depends on
a) current through material
b voltage of circuit
6 nature of material
(1) barometric pressure 4) The power factor, a) is a measure of how much energy the electrons contain while moving along a
conductor. b) has units called vomps and is the power in watts multiplied by the resistance in
ohms.
is less than 1.0 if more current is ﬂowing to the load than is required to supply the
Aim actual power used by the load.
A cappedyd) is a measure of efﬁciency and is the output power divided by the input power. 5) The purpose of giving CPR to an unconscious shock victim is to
a) revive the person. b) prevent stoppage of the heart.
c prevent ventricular ﬁbrillation ﬁ'om occurring.
h @ keep person alive until deﬁbrillation assistance can arrive. Page 1 of7 6) Negative instantaneous power in an AC circuit
@ means energy stored in a reactive load component is ﬂowing ﬁom the load to the
generator.
b) is impossible.
c) means that resistive load components are adjusted to lower voltage by passing
stored energy back to the generator.
(1) occurs only at power factors less than 0.5. 7) A 600  Watt, lZOvolt air conditioner is operated 8 hours each day. If electrical
energy costs 5 cents per kWh, how much will the operation of this load add to the bill for a 30—day billing period? ' > , .
$7.20 4965 c) $7,200 3 2 Mil Idiot»
‘1) $60 qu 005 ) $2.40 = i. 20 f) none of the abOVe ‘W a 8) If 60 Hz shock current is gradually increased, muscular control will be lost by the
time this level of current is reached: 301m <Q‘ZSMA WWI Todth 3.1
b 6mA @16m/ 66% qBOMA run/w Pang
d)160mA e) 1600 9) How long will it take a 4kW load to use 12 kWh of energy?
3 hour b) 1/3 hour ll in
c) 36 minutes = 3k
d) 3.6 hours e) none of the above 10) A l20volt appliance has a power factor of 0.6 and uses 850 watts. The current
flowing to the appliance is: 11.8 amps
) 7.08 amps
c) 4.25 amps d) 0.235 amps
e) none of the above Page 2 of 7 11) The purpose of the National Electrical Code is to
@ safeguard persons, buildings, and contents ﬁom electrical hazards.
b) insure that electrical systems are economical and efﬁcient.
c) insure that electrical systems have adequate capacity
(1) a & b
e) b & c
t) a & c 12) When an extra parallel load is added to a parallel circuit, total resistance of the circuit:
a) goes up.
goes down.
c) is not changed
d) can go up or down depending on circuit voltage. 13) Most electrocutions on American farms occur:
a) because of electrical equipment failure.
b) because of lightning.
c) while the person is doing wiring work.
@ as a result of contacting high voltage overhead lines. 14) A 3wire 120/240 volt branch circuit has only 120volt resistive loads connected.
Currents measured: Black: 10 amps White: 12 amps The wattage of the load connected between red and white is: __ a) 1440 watts f) Q 20 V 1 21A)  F b) 1200 watts ‘ I _ C 240watts R W” P
2640 watts e can't tell from information given 15) Severity of an electrical shock depends on
a) magnitude of current.
b) duration of shock.
c route through body.
all of the above.
e) a & b
t) a & c 16) A voltmeter placed in parallel with a load will:
a) cause a short circuit, but no damage to the meter.
@ read correctly the voltage across the load.
c) shut off almost all current to the load.
d) cause a short circuit and ruin the meter. Page 3 of 7 17) A watthour meter has a Kh factor of 3 on the nameplate. With a test load connected
through it, 10 revolutions of the disk take 300 seconds. Compute the load wattage. / 360 r . ,
120 ‘0 V94“ 300, 36005 = 36 O V
c) 9000 watts 300 )3?" V520 l bk d) 90 watts
e) none of the above 18) A threeway switch (S3) is actually a: a SPST
é SPDT
c) DPST
d) DPDT 19) The sum of the voltage drops across the loads in a series circuit is equal to a) the current squared.
b) the total power dissipated.
c) the current divided by the total resistance. the applied voltage. 20) Name two major advantages of connecting electrical loads in p b) ° bum. resisted/nu, ° high,“ 0W ‘chn/O ° €W&% addl MOLL loam/db) lad—exp. arallel rather than Page 4 of7 Part 2 — Formula Sheet (one side 8.5”x11”)
Long Essay: 20 points each (60%)
SHOW YOUR WORK (no work = no credit) 21) Use the following seriesparallel circuit diagram to answer parts ad.
£2 ISIL Boil
240V
5A 20.57.
I l
5?}; a) Compute RT #— 2 i— + ‘—
RT' £1: 230V 1 LlCZJl"2T R23 R2 R3
" IT 5A 2 7i + .\_
3075 so
5 ﬁg b) ComputeR4 2 239 : ‘OJ
I) RT=2.+RZ?>+Rq~+RS 7‘3 3
‘18 = 10+ Io + 2% +20
LHZ=Rq+qo Razgﬂ
5 k c) Compute 13
F E23 : IT [223 V: ‘7‘ I :2 E23 _: —_ I v; I
3 23 BOIL / 3
5 F); (1) Compute Power used by R;
E4 : IT 2.4 = (5A)(<UL) = 40V
‘ (,3 = IT EL, = (SAYqov) ~= now = E.
Page50f7 22) Use the following 120/240V 3wire circuit diagram to answer parts ab ’(0P+Sa) Compute 11 ‘= 3A 12=3A
13 =(pA
I4 =?A
15 =(pA
16 =HA
I7 =A
IzzillA ‘1 ﬁsh) Compute Total Power used by the circuit
P  (zqov\£3A) = 3‘20“) P : (\wVXH/O ‘= 2(3qu Page 6 of 7 23) Use the following R—C circuit diagram to answer parts a—e. 2 = 8n.
VLOV
éOHa
xcr— an
ComputeC
C erF C 211+ Xc 21r( GO HLYQJL)
CL = 0.0004411 F 2 “Hal F
Compute Z and express in polar and rectangular notation.
‘Z‘ = (2+3 (xUXL‘) = $43 (—a) = gem6.: Z
9 .
wig: ‘W= c = lo a; : ,OL—amr
_ _‘ 9.: mm" (A?) = 3M?“
Compute I and express in polar and rectangular notation. (Hint: Reference I) x i
 c _ no [.92 '—' 3%”
I o L — 36. 23‘
0 = 92 “we? 92 = GM?”
Sketch a phasor diagram referencing I and showing ET, ER, Ec, and phase angle. _,/ 6! (0V (Include magnitudes) ET=120 1.36.87?” __ E2 = no Cea('36.(67‘) = 95.3 =Cl(o\/
EC = :20 gm (aw?) = —?2.ov Compute Reactive Power PR P2 = IE gm 9 : (l‘ZAXlloVXSIwBBQRﬁ) PK 2 VOA/Oak Page 7 of 7 ...
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 Electrical Engineering, Volt, extra series load

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