Compare an RC circuit with an LR circuit, The picture to the right shows a circu
ID: 2111205 • Letter: C
Question
Compare an RC circuit with an LR circuit, The picture to the right shows a circuit with a battery, a capacitor, a resistor, and a switch. Suppose we initially have a situation where the switch has been in position 1 for a long time and then is moved to position 2 at t = 0. Make three graphs: i) (Delta V compactor + Delta V resistor) vs t, ii) Delta F capacitor vs t, and iii) Delta V resistor vs t for times while the switch is in position 2. Include some time before t = 0 as well as after so that we can see the change at t = 0. After a long time we now move the switch back to position 1 at time t = t1. Make three more graphs: i) (Delta V capadtor + Delta V resistor) vs t, ii) Delta V capadtor vs t, and iii) Delta V resistor vs t for times while the switch is in position 1. Include a little time before t = t1 as well as after so that we can see the change at t = t1. The picture to the right shows a circuit with a battery, an inductor, a resistor, and two switches. Suppose we initially have a situation where switch S2 has been closed for a long time while switch Si has been open. At time t = 0, we open S2 at the same time that we close Si. Make three graphs: i) (Delta inductor + Delta V resistor) vs t, ii) A inductor vs t, and iii) AV resistor vs t for times while S1 is closed. Include some time before t = 0 as well as after so that we can see the change at t = 0. After a long time we open S1 and simultaneously close S2 at time t = t1. Make three graphs: i) (Delta Inductor + Delta V resistor) vs t, ii) Delta Inductor vs t, and iii) AV resistor vs t for times while S2 is closed. Include some time before t = ti as well as after so that we can see the change at t = t1.Explanation / Answer
a) at t=0 , capacitor behaves as short circuit so no charge will pass through it as time passes charge will flow through it at a rate of q= qo(1-e^-t/RC) and V=Vo(1-e^-t/RC)and at t= inifite it becomes fully charge so all chrge will flow through it and it behaves as infinite resistance wire....
and when switch goes to position 2 capacitor will dipites the charge wrt to time with relation q=qo*e^-t/RC and V=Vo*e^-t/RC
from both the relation of voltages and charge we get the graphs....
b) but inductance behaves just opposite to capacitance at t=0 , it behaves as infinite resistance wire and all the current will pass through it with rate of i= io(1-e^tR/L)...
and after a long time at t= infite it behaves like short circuit and no currnet will pass through it....
and decay currnt is i=io*(e^-tL/R).
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