152 Direct Current Circuits*P28.62(a)∆∆VVIRIRII I IIRRIRRRIIRIRIRI1211221211121212122212===+=+=+=++=IR1R2I1I2FIG. P28.62(a)(b)The power delivered to the pair is P−IR IRI I R122222bg. For minimum powerwe want to find I1such that ddI10=.ddII IR1122210=+−−=af20−+=IIR12=+This is the same condition as that found in part (a).P28.63Let Rm=measured value, R=actual value,IR=current through the resistor RI=current measured by the ammeter.(a)When using circuit (a), V−∆20 000orRIIR=−LNMOQP20 0001 .But sinceIVRm=∆and IVRR=∆, we haveIIRRRm=(a)(b)FIG. P28.63andRRmm=−20 000.(1)When
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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 .