RC circuit graphs

[chiddler]

answer is D. My question is what happens if you add another capacitor in series? Would the graph just shrink? (slope becomes more negative?)

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[milski]

Adding another capacitor in series will decrease the equivalent capacitance. The starting current is E/R and will stay the same. The decay is e^(-t/RC). If you decrease C, t/RC will increase and -t/RC will decrease. The graph will get squished towards the horizontal axis and will approach it faster. It's the opposite of what happens when you add a capacitor in parallel. Note that the graph will not exactly shrink, in a sense that the point where it crosses the y axis will stay the same.

 

[chiddler]

Adding another capacitor in series will decrease the equivalent capacitance. The starting current is E/R and will stay the same. The decay is e^(-t/RC). If you decrease C, t/RC will increase and -t/RC will decrease. The graph will get squished towards the horizontal axis and will approach it faster. It's the opposite of what happens when you add a capacitor in parallel. Note that the graph will not exactly shrink, in a sense that the point where it crosses the y axis will stay the same.

right. it'll reach max capacitance and current = 0 faster.

thank you for answering.