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34.Indakan.The inductor in a radio receiver carries a current of amplitude183 mAwhenit is connected to a voltage of amplitude2.45 Vand frequency1.36 MHz.What is theinductance?A.1.57μHB.3.13μHC.8.74μHD.9.84μH35.PHASE-OR-DIE-A-GRAM.A LRC series circuit, operated by an AC source, has cur-rentI, inductor voltageVL, capacitor voltageVC, and resistor voltageVR. Which of thefollowing is a correct phasor diagram of the circuit?
Second Long ExamSecond Semester, AY 2016–2017Physics 7236.Arremess-ing.What is the root-mean-square value of this sinusoidal voltage?37.sad reacts only.At a certain angular frequencyωof an AC current, the reactances of a4-mHinductor and a1-mFcapacitor are equal. What is this reactance?38.Constantly Difficult.An LRC series circuit is operated by an ACsource. Which of the following combinations ofR,XLandXCwill result to the phasor diagram of the voltage and the currentthrough the circuit shown in the figure?ChoicesR[Ω]XL[Ω]XC[Ω]A3.54.52B85.24.3C22.214.171.124D126.96.36.1999.ImPhiDance.Consider the phasor diagram of an LRCseries circuit showing the current amplitudeI, resistorvoltageVR, source voltageVS, and the phase angleφ.What is the impedance of the circuit?8 ΩA – 12
Second Long ExamSecond Semester, AY 2016–2017Physics 7240.Optimus Secondary.A transformer connected to a180-V(rms) ac line is to supply15.0 V(rms) to a portable electronic device. Which among the following pairs of primary andsecondary turn count of the transformer best achieves this goal?A – 13
Second Long ExamSecond Semester, AY 2016–2017Physics 72A – 14
Second Long ExamSecond Semester, AY 2016–2017Physics 72A – 15
Second Long ExamSecond Semester, AY 2016–2017Physics 72PrefixAbbreviationFactorteraT1012gigaG109megaM106kilok103centic10-2millim10-3microμ10-6nanon10-9picop10-120π/6π/4π/3π/2sin01/21/√2√3/21cos0√3/21/√21/20tan01/√31√3undefk= 8.988×109N·m2/C2ε0= 8.852×10-12C2/N·m2|e|= 1.602×10-19Cmelectron= 9.11×10-31kgmproton= 1.672×10-27kgμ0= 4π×10-7T·m/Ac= 2.998×108m/sS= 4πr2V=43πr3S= 2πrh+ 2πr2V=πr2h~F=q~E~F=q~v×~B;~F=I~l×~BΦB=~B·~AI~B·d~A= 0~B=μ04πq~v×ˆrr2d~B=μ0I4πd~l×ˆrr2B=μ0I2πr;B=μ0I2R2(R2+d2)3/2I~B·d~l=μ0If=μ02πI1I2rE=I~E·d~l=-dΦBdtE=Z~v×~B·d~liD=εdΦEdtE=-MdIdt;E=-LdIdtuB=B22μ0U=Q22C;U=12LI2ω=1√LCR2>4L/C;R2= 4L/C;R2<4L/CXL=ωL;XC=1ωCtanφ=XL-XCRZ=qR2+ (XL-XC)2V1V2=I2I1=N1N2A – 16