Initially the circuit is in steady state with switch closed. Now at t = 0 the switch is opened. The voltage across the capacitor V(t) for t >0 is given by ${\text{V (t) = A + B e}}^{\text{-4t}}\text{ Volts}$ . Find the ratio $\frac{\text{A}}{\text{B}}$.

At the instant just before t = 0, the equivalent circuit is shown in figure below
  
 $I=\frac{40}{10+5+\frac{10\times 10}{10+10}}=2A$                  $\therefore V_C=V_A-V_B=5\times 2=10V$   
 At time  $t\to \infty$, the equivalent circuit is shown in figure below : 

 $I=\frac{40}{10+5+10}=\frac{40}{25}$    $\therefore V_C=V_A-V_B=5\times \frac{40}{25}=8V$  
 The equivalent resistance across the capacitor is given by 
$R_{eq}=\frac{(10+10)5}{20+5}=4\Omega$                $\therefore \tau =$ Time constant of the discharge circuit
$=C{R}_{eq} =\left(\frac{1}{16}\right)\left(4\right)=\frac{1}{4}\sec$.        Now, $V_C=A+B{e}^{-kt}$ 
 At t = 0, $V_C=10V$  or, 10 = A + B
 At  $t\to \infty ,V_C=8V$ or, 8 = A + 0  or,  A = 8V
      B = 2V Also , $k=\frac{1}{\tau }=\frac{1}{1/4}=4s^{-1}$