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Introductory Physics Homework Help
Graphing Current v. Time for simple 2R1C circuit
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[QUOTE="friendbobbiny, post: 4991079, member: 506810"] 1. Homework Statement [ATTACH=full]175740[/ATTACH] The switch is closed. Graph the current in [itex]R_{1}[/itex] 2. Homework Equations [itex] V = IR [/itex] [itex] CV = Q [/itex] 3. The Attempt at a Solution 1) [itex]R_{0}[/itex] can be ignored a) The resistor simply reduces the voltage before any current can be split. b) We're justified, as a result, in imagining that [itex]R_{0}[/itex] were removed.2) Through the branch with C and [itex]R_{2}[/itex], current will eventually stop flowing a) Capacitor C will acquire a potential difference of V b) When that happens, Kirchoff's loop rule -- that voltage drops and gains belonging to a circuit LOOP sum to zero -- won't need to be fulfilled by having current pass through [itex]R_{2}[/itex] c) When current doesn't pass through [itex]R_{2}[/itex], the branch can't be completed and current won't flow through it at all [I][B] d) Is this sound reasoning? Or, is it the case that current does continue flowing through the capacitor but just won't go any further? [/B][/I] 3) If current doesn't enter the capacitor, more of it enters [B][itex]R_{1}[/itex],[/B] our resistor of interest [B] [/B]a) The current entering [B][itex]R_{1}[/itex] [/B]will change at the rate that current decreases the capacitor C. b) Eventually, current approaches an asymptote: V/R4) The resulting graph is logarithmic, [I]except that it has an asymptote at y = [itex]V/R[/itex] Note! Do not rely on this graph. Just frames the idea.[/I] [ATTACH=full]175741[/ATTACH] [/QUOTE]
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Introductory Physics Homework Help
Graphing Current v. Time for simple 2R1C circuit
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