Because the current passing through R results in a voltage drop across R.*Why is it that the graph approaches an asymptote of vc, and not +9V?
wassabi said:Because the current passing through R results in a voltage drop across R.*Why is it that the graph approaches an asymptote of vc, and not +9V?
9V - VdropR = Vcap
wassabi said:Rephrasing: the current passing through R will decrease as C approaches 100% charge. The Vdrop across R then approaches but never actually hits zero.
stefannm said:2. I don't know. By the definition of tangent line, you could draw it anywhere on the graph. Perhaps it can be argued that the origin is the most interesting because it will give an indication of the maximum current because i = C dv/dt where dv/dt is the slope of the tangent line.
stefannm said:Thanks LvW. I never realized that before.
wassabi said:Rephrasing: the current passing through R will decrease as C approaches 100% charge. The Vdrop across R then approaches but never actually hits zero.
stefannm said:1. The asymptote should be 9V. I think the graph is off by a factor of 10.
LvW said:The graph is right, it the memo I got, .......
Ahh - I see. OK, a memo can't be wrong.
In this case, we have to rewrite all of the textbooks.
matbob said:No, the asymptote would be at 9V. Let the capacitor charge to 9V which happens at t = infinity. Now what the resistor sees is 9V on both its terminals. So there won't be any current through the resistor which means no drop across it.
matbob said:When the capacitor is fully charged, ie, when it is at 9V, there won't be any current flow (provided, the capacitor is loseless). So there won't be a pot difference across the resistor.
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