Chapter 27127*P27.75(a)Think of the device as two capacitors in parallel. The one on the left has κ11=,Ax12=+FHGIKJAA. The equivalent capacitance isκκ10 120 2000222∈+∈=∈+FHGIKJ+∈−FHGIKJ=∈++−AdAddxdxdxxAAAAAAAaf.(b)The charge on the capacitor is QCV=∆QVd=∈02A∆af.The current isIdQdtdQdxdxdtVdvVvd===∈=−∈−00202021∆∆afaf.The negative value indicates that the current drains charge from the capacitor. Positivecurrent is clockwise ∈−01A∆Vvd.ANSWERS TO EVEN PROBLEMSP27.2364. hP27.32171ΩP27.4(a) see the solution; (b) 105mAP27.340153ΩP27.6(a) 17 0. A ; (b) 85 0. kAm2P27.36500A, 24 0ΩP27.38448 AP27.8(a) 99 52; (b) 800mmP27.40(a) 0.530; (b) 221 J; (c) 15.1°CP27.10(a) 221 nm ; (b) no; see the solutionP27.42(a) 317m ; (b) 340 WP27.1230 3. MAm2P27.44(a) 0 660.kWh ; (b) 3 96¢.P27.14(a) 3.75 kΩ; (b) 536 mP27.46(a) 2.05 W; (b) 3.41 W; no
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .