Consider a cirucuit made of two resistors of resistances r and R and a capacitor

Question

Consider a cirucuit made of two resistors of resistances r and R and a capacitor of capacitance C attached to a battery of emf E (where the internal resistance is included in r) as in the figure. Initially the switch is open and the capacitor is empty.

Denote the current running through resistance r (from left to right) to be Ir, the current running through resistance R (from top to bottom) to be IR, and the charge that accumulates on the capacitor to be Q(> 0). (They are all time-dependent quantities.)

1. Using the loop rule applied to the left loop (consisting of E, r and C), express the relation between Ir and Q .

2. Using the loop rule applied to the right loop (consisting of R and C), express the relation between IR and Q .

3. Using the junction rule, write dQ/dt in terms of Ir and/or IR.

4. Combining 1, 2, and 3 above, write a differential equation for Q(t) in the form

where a and b are constants. Identify a and b in terms of E, r, R, and C.

5. Using the fact that the above dierential equation is solved to be

write out Q(t) in terms of E, r, R, and C and arbitrary constant A.

6.Determine A from initial condition Q(0) = 0. (Switch is closed at t = 0).

Explanation / Answer

1) Applying loop rule in left loop, starting from a point below battery, ( using potential dorp across resistor = ir and across capacitor = q/C) we get
E - Ir (r) - Q / C = 0

2) Starting above the calapcitor and moving in the clockwise direction we get
- IR (R) + Q/C = 0

3) Current coming to the upper junction = current leaving it.
Ir = IR + dQ/dt ( current through capacitor = rate of change of charge)

4) Eliminating IR from equation 2 and 3 we get
Ir = Q/RC + dQ/dt
Substituting this in equation 1 , we get
E - r Q/RC - r dQ/dt - Q/C = 0
dQ/dt + Q/C ( 1/R + 1/r) = E/r

5) Writing 1/Re = 1/R + 1/r
Last equation is
dQ/dt + Q / Re C = E/r
Rearranging terms we get
dQ/ (E/r - Q/ ReC) = dt
Integrating this we get
- Re C Ln( E/r - Q/Re C) = t + K (K is cosnatnt of integration)
Q = E Re C / r + A e-t/ReC

6) Using condition t= 0 , Q = 0 , we get
A = - E Re C / r

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