Question

In the figure shown , the capacitor is initially uncharged. The current in resistor $R_3$ at time $t$ is $\frac{E}{2R}(1-e^{\frac{-At}{BRC}})$. Consider that $R_1=R_2=R_3=R$, the value of $A+B$ will be:

Answer 1

Showing the current distribution at any time $t$ in the circuit.


Applying $\text{KVL}$ in loop 1 and loop 2:

$E-(i_1+i_2)R-i_{1}R=0...\left(1\right)$

$-i_{2}R+E-\frac{q}{C}+i_{1}R=0...\left(2\right)$

Multiplying Equ (2) by $2$ and adding with Equ (1)

$\Rightarrow 3E-3i_{2}R-\frac{2q}{C}=0$

$\Rightarrow \frac{3E}{2}-\frac{q}{C}-\frac{3}{2}{i}_{2}R=0$

Substituting $i_2=\frac{dq}{dt}$ we get;

$\frac{3CE-2q}{2C}=\frac{3}{2}R\left(\frac{dq}{dt}\right)$

$\Rightarrow \frac{dq}{3CE-2q}=\frac{dt}{3RC}$

Integrating both sides,

${\int }_0^q\frac{dq}{3CE-2q}={\int }_0^t\frac{dt}{3RC}$

$\Rightarrow \frac{\left[\ln \right(3CE-2q\left)\right]{|}_0^q}{-2}=\frac{t}{3RC}$

$\Rightarrow \frac{-1}{2}\left[\ln \right[3CE-2q]-\ln [3CE]=\frac{t}{3RC}$

$\Rightarrow \frac{-1}{2}\ln \left[\frac{3CE-2q}{3CE}\right]=\frac{t}{3RC}$

$\Rightarrow 1-\frac{2q}{3CE}=e^{\frac{-2t}{3RC}}$

$\Rightarrow q=[1-e^{\frac{-2t}{3RC}}]\frac{3CE}{2}$

$\Rightarrow i_2=\frac{dq}{dt}=\left(\frac{-3CE}{2}{e}^{\frac{-2t}{3RC}}\right)\times \left(\frac{-2}{3RC}\right)$

$\Rightarrow i_2=\left(\frac{E}{R}\right)e^{\frac{-2t}{3RC}}$

From equ (1),

$i_1=\frac{E-i_{2}R}{2R}=\frac{E-E{e}^{\frac{-2t}{3Rc}}}{2R}$

$\Rightarrow i_1=\frac{E}{2R}[1-e^{\frac{-2t}{3RC}}]$

Comparing the relation with $\frac{E}{2R}(1-e^{\frac{-At}{BRC}})$ we get,

$A=2$ and $B=3$

$\Rightarrow A+B=5$

Accepted answer : $5$.