Initially switch is open, no charge on the capacitor, (a) Close the switch, find I_i, (i = 1, 2, 3) Q, & V_C immediately after. (b) Switch is closed for long find I_i, Q & V_C. (c) Find I_i, Q, & V_c immediately reopened? d) Reopened for long?

Feb 21, 2017

See below.

Explanation:

Initially switch is open, no charge on the capacitor, (a) Close the switch, find I_i, (i = 1, 2, 3) Q, & V_C immediately after. (b) Switch is closed for long find I_i, Q & V_C. (c) Find I_i, Q, & V_c immediately reopened? d) Reopened for long?

Calling ${R}_{1} = 12.0 \Omega$, ${R}_{3} = 3.00 \Omega$, $E = 9.00 V$ and $C = 10.0 \mu F$
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 $Q = {\int}_{0}^{t} I \mathrm{dt}$ 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 ${\dot{Q}}_{1}$ and ${\dot{Q}}_{2}$ 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
${\alpha}_{1} = \frac{{R}_{2} - {R}_{3}}{{R}_{1} {R}_{2} + {R}_{2} {R}_{3}}$ and
${\alpha}_{2} = \frac{{R}_{1} + {R}_{2}}{C \left({R}_{1} {R}_{2} + {R}_{2} {R}_{3}\right)}$
${\beta}_{1} = \frac{1}{{R}_{1} + {R}_{3}}$ and
${\beta}_{2} = \frac{1}{C \left({R}_{1} + {R}_{3}\right)}$ 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.

(a) At the very init ${Q}_{2} \left(0\right) = 0$ so we have
${c}_{2} + \frac{{\beta}_{1}}{{\beta}_{2}} E = 0$ so ${c}_{2} = - \frac{{\beta}_{1}}{{\beta}_{2}} E$
also
${Q}_{1} \left(0\right) = 0$ so we have

${c}_{2} + {c}_{1} = 0$ then ${c}_{1} = \frac{{\beta}_{1}}{{\beta}_{2}} E$ and the charge equations are

${Q}_{1} = - \frac{{\beta}_{1}}{{\beta}_{2}} E \left(1 - {e}^{- {\beta}_{2} t}\right)$ and
${Q}_{2} = \frac{{\beta}_{1}}{{\beta}_{2}} E \left(1 - {e}^{- {\beta}_{2} t}\right)$

Analogously

${I}_{1} = {\dot{Q}}_{1} = - {\beta}_{1} E {e}^{- {\beta}_{2} t}$ and
${I}_{2} = {\dot{Q}}_{2} = {\beta}_{1} E {e}^{- {\beta}_{2} t}$