Initially switch is open, no charge on the capacitor, (a) Close the switch, find I_i, (i = 1, 2, 3) Q, & V_CIi,(i=1,2,3)Q,&VC immediately after. (b) Switch is closed for long find I_i, Q & V_CIi,Q&VC. (c) Find I_i, Q, & V_cIi,Q,&Vc immediately reopened? d) Reopened for long?
![enter image source here]()
Calling R_1=12.0Omega, R_3=3.00Omega, E = 9.00V and C = 10.0muF
After closing the switch we have
{(E = I_1 R_1+(I_1-I_2)R_2),
(1/C int_0^tI_2dt +I_2 R_3+(I_2-I_1)R_2=0)
:}
but Q = int_0^t I dt so the equations can be stated as
{
(E = dotQ_1 R_1+(dotQ_1-dotQ_2)R_2),
(1/C Q_2 +dotQ_2 R_3+(dotQ_2-dotQ_1)R_2=0)
:}
Solving for dotQ_1 and dotQ_2 we get
{
(dotQ_1 =(C E- Q_2)/(C (R_1 + R_3))),
(dotQ_2 = (C E (R_2-R_3) - Q_2 (R_1+R_2))/(C (R_1 R_2 + R_2 R_3))):}
Calling
alpha_1=(R_2-R_3)/ (R_1 R_2 + R_2 R_3) and
alpha_2 = (R_1+R_2)/(C (R_1 R_2 + R_2 R_3))
beta_1=1/(R_1+R_3) and
beta_2=1/(C(R_1+R_3)) we have
{
(dotQ_1+beta_2Q_2=beta_1E),
(dotQ_2+alpha_2Q_2=alpha_1E):}
Solving the differential equations we have
{
(Q_2(t)=c_2e^(-beta_2 t)+(beta_1)/(beta_2)E),
(Q_1(t)=c_2e^(-beta_2 t)+c_1):}
where c_1 and c_2 are integration constants to be defined according to the initial conditions.
Answering the required items
(a) At the very init Q_2(0)=0 so we have
c_2+(beta_1)/(beta_2)E=0 so c_2=-(beta_1)/(beta_2)E
also
Q_1(0)=0 so we have
c_2+c_1=0 then c_1=(beta_1)/(beta_2)E and the charge equations are
Q_1=-(beta_1)/(beta_2)E(1-e^(-beta_2t)) and
Q_2=(beta_1)/(beta_2)E(1-e^(-beta_2t))
Analogously
I_1= dot Q_1=-beta_1Ee^(-beta_2t) and
I_2=dot Q_2=beta_1Ee^(-beta_2t)
