Direct-Current Circuits129±±²³³´µ¶·²³´²±¶±·±¸¸²³³¹²¶³²³³º²¶»²³¹²)18.49(a) From (loadIr Rε=+, the current supplied when the headlights are the entire load is ()12.6 V2.48 A0.080+5.00 loadIrR===+ΩThe potential difference across the headlights is then 2.48 A5.00 12.4 VloadVIR∆==Ω=(b) The starter motor connects in parallel with the headlights. If is the current supplied to the headlights, the total current delivered by the battery is The terminal potential difference of the battery is hlIIr35.0 AhlIIV∆=−, so the total current is ( )IVr∆while the current to the headlights is 5.00 hl=∆Ω. Thus, becomes 35.0 Ahl35.0 A0 VV
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.