The circuit in the figure has been in position a for a long time, then the switch is thrown to position b. With #Vb = 12 V, C = 10 mF, R = 20 W#. a.) What is the current through the resistor before/after the switch? b) capacitor before/after c) at t=3sec?

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1 Answer
Mar 24, 2017

Answer:

See below

Explanation:

[NB check units of resistor in question, assume it should be in #Omega#'s]

With the switch in position a, as soon as the circuit is complete, we expect current to flow until such time as the capacitor is charged to the source's #V_B#.

During the charging process, we have from Kirchoff's loop rule:

#V_B - V_R - V_C = 0#, where #V_C# is the drop across the capacitor's plates,

Or:

#V_B - i R - Q/C = 0#

We can differentiate that wrt time:

#implies 0 - (di)/(dt) R - i/C = 0#, noting that #i = (dQ)/(dt)#

This separates and solves, with IV #i(0) = (V_B)/R#, as:

#int_( (V_B)/R)^(i(t)) 1/i (di)/(dt) \ dt = - 1/(RC) int_0^t \ dt#

#i = (V_B)/R e^(- 1/(RC) t)#, which is exponential decay .... the capacitor gradually charges so that the potential drop across its plates is equal to source #V_B#.

So, if the circuit has been closed at a for a long time, then #i = 0#. So no current through either the capacitor or resistor before the switch to b.

After the switch to b , we are looking at an RC circuit, with the capacitor discharging to the point the drop across its plates is zero.

During the discharging process, we have from Kirchoff's loop rule:

#V_R - V_C = 0 implies i R = Q/C#

Note that, in the discharge process: #i = color(red)(-) (dQ)/(dt)#

Again we can differentiate that wrt time:

# implies (di)/(dt) R = - i/C#

This separates and solves as:

#int_(i(0))^(i(t)) 1/i (di)/(dt) \ dt = - 1/(RC) int_0^t \ dt#

#implies i = i(0) e^(- t/(RC))#

In this instance, because the capacitor is fully charged and so has voltage #V_B#, we know that #i(0) = V_B/R = 12/20 = 0.6A#.

That is the current immediately the switch is closed at b.

And so:

# i(t) = 0.6 e^(- t/(RC))#

Finally at #t = 3# we have :

# i(3) = 0.6 e^(- 3/(20 cdot 10^(-2)) ) = 1.8 times 10^(-7) A#