31_Chapter 18 HomeworkCH18 Direct-Current Circuits

31_Chapter 18 HomeworkCH18 Direct-Current Circuits -...

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Direct-Current Circuits 129 ± ±²³³´µ ¶·²³´ ² ± ± · ± ¸ ¸²³³ ¹ ² ¶³²³ ³ º ² ¶»²³ ¹ ² ) 18.49 (a) From ( load Ir R ε =+ , the current supplied when the headlights are the entire load is () 12.6 V 2.48 A 0.080+5.00 load I rR == = +Ω The potential difference across the headlights is then 2.48 A 5.00 12.4 V load VI R ∆= = 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 hl I I r 35.0 A hl II V =− , so the total current is ( ) IV r while the current to the headlights is 5.00 hl = ∆Ω . Thus, becomes 35.0 A hl 35.0 A 0 VV
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