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Electronic circuits and components * Course Index * Introduction * About the Author * About this Course * Feedback * Course Navigation * How to use this Course * TINA * Locktronics Fundamentals * Introduction: Fundamentals * Units and Multiples * Electricity * Electronic Principles * Electrical Circuits ## Basic Lamp Circuit ## Lamps in Series ## Lamps in Parallel * Alternating Current * Assessment: Fundamentals Passive Components * Introduction: Passive Components * Resistors * Capacitors * Inductors * Transformers * Batteries, Fuses, Lamps and Switches * Assessment: Passive Components Semiconductors * Introduction: Semiconductors * Diodes * Transistors * Logic Gates * Assessment: Semiconductors Passive Circuits * Introduction: Passive Circuits * Series and Parallel Connections * Kirchoff's Laws * Potential and Current Dividers * Passive Time Variant Circuits * Assessment: Passive Circuits Active Circuits * Introduction: Active Circuits * Power Supply Circuits * Operational Amplifier Circuits * Transistor Amplifier Circuits * 555 Timer Circuits * Assessment: Active Circuits Parts Gallery * Introduction: Parts Gallery * Passive Component Images * Semiconductor Images * Other Images * Quizzes |
(:Summary:Contains the 'action' links (like Browse, Edit, History, etc.), placed at the top of the page, see site page actions:) (:comment This page can be somewhat complex to figure out the first time you see it. Its contents are documented at PmWiki.SitePageActions if you need help. :) * Print (:comment (:if group Site,SiteAdmin,Cookbook,Profiles,PmWiki*:) (:comment delete if and ifend to enable backlinks:) * %item rel=nofollow class=backlinks accesskey='$[ak_backlinks]'% [[{*$Name}?action=search&q=link={*$FullName} | $[Backlinks] ]] (:ifend:) :) * Login Lamps in Parallel<^< Lamps in Series | Course Index | Alternating Current >^>(:nl:) ![]() In the parallel connected circuit, two 6V 3W lamps are connected in parallel in order to consume a total power of 6W from the 6V supply. Once again, the current flowing in each lamp can be calculated from: I = P/V = 6W/3V = 0.5A Note that the total current delivered by the battery must be twice this value, or 1A. Applying Ohm's Law to determine the working resistance of each lamp (i.e. the resistance when illuminated) gives: R = V/I = 6V/0.5A = 12Ω Since the circuit demands a total current of 1A (i.e. 0.5A + 0.5A), the effective resistance of the circuit must once again be 6Ω. (:nl:)(:table style="clear:both":)
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