Thanks,, 3. Consider the circuit below. Initially both capacitors are uncharged.
ID: 1581755 • Letter: T
Question
Thanks,, 3. Consider the circuit below. Initially both capacitors are uncharged. Then the switch, which was initially open is move to position b and left for a long time. Then at t = 0 (the began at t =-oo), the switch is moved to positio capacitors at t = 0 versus a long time after? (b) as a function of t for t>0 t>0 n a. (a) What is the energy stored in the Find the current moving through resistor R (c) Find the energy delivered to the resistors as a function of t for as a function of t for t 0 tExplanation / Answer
Initially when you throw switch to b position ,current starts flowing through R and C1,till C1 is fully charged. The capacitor is charged fully and has energy of 0.5 C1 E^2.
Thus immediately when swich is thrown towards position a , The energy is;
In C1 = 0.5 CE^2 and in C2 ,the energy = 0
After a long time,
The charge redistributes between the two capacitors ,which are now in Parallel
Both capacitors will have a final voltage Vf common to both
Let combined final charge in both capacitors in parallel be Qf,
Qf of the two capacitor system = Q1f + Q2f = Q0 ( Q0 being the initial charge on C1)
C1E = C1 Vf + C2 Vf ( Vf being final equilibrium voltage after which no charge redistribution takes place)
Vf = C1 E /(C1+C2)
Energy sored in C1 = 0.5 C1 [C1E/(C1+C2)]^2 = 0.5 C1^3 X E^2/( C1+C2)^2
Energy stored in C2 = 0.5 C2 X C1 [C1E/(C1+C2)]^2 = 0.5 C1^2 C2 E^2 /(C1+C2)^2
Total Energy stored = 0.5 X [ C1^2E^2/(C1+C2)^2] [ C1+C2] = 0.5 C1^2E^2 / ( C1+C2)
(b) Current flow as function of time for t>0 is required
The current starts flowing such that voltage between two terminals of resitor
V(t) = The voltage across the resistor = Ee(-t/RC) , since V=E at t=t and V=0 at t= infinity
current = - (E//R) e^(-t/RC)
where C = C1C2/C1+C2
(c) Power = (E^2/R^2) XR X e^(-2t/RC) since P = i^2 R
=(E^2/R) X e^(-2t/RC)
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