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A pure induct...

Question

A pure inductor, pure capacitor, and two pure equal resistors are connected as shown in the circuit, such that $L = 2 R^{2} C$ .
At an instant $t = 0$ , the switch is closed. If $i_{1}$ and $i_{2}$ are currents shown in branches at some time $t = t_{o}$ relation $\frac{d i_{1}}{d t} + \frac{d i_{2}}{d t} = 0$ is satisfied, then $t_{o} = n R C$ . Here $n$ is
(Take $ln 2 = 0.69$ and answer up to two decimal places)

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Solution

$i_{1} = \frac{ε}{R} [1 - e^{- R t / L}]$
$i_{2} = \frac{d q}{d t} = \frac{d}{d t} [q_{o} (1 - e^{- t / R C})] i_{2} = \frac{ε}{R} e^{- t / R C}$
So,
$\frac{d i_{2}}{d t} = - \frac{ε}{R^{2} C} e^{- t / R C}, \frac{d i_{1}}{d t} = \frac{ε}{L} e^{- R t / L}$
$\frac{d i_{1}}{d t} + \frac{d i_{2}}{d t} = \frac{ε}{L} e^{- R t / L} - \frac{2 ε}{L} e^{- 2 R t / L} = 0$
$\Rightarrow 2 e^{R t_{o} / L} = e^{2 R t_{o} / L} \Rightarrow ln 2 + \frac{R t_{o}}{L} = \frac{2 R t_{o}}{L}$
$\Rightarrow t_{o} = 2 (R C) ln 2$
$\Rightarrow n = 2 ln 2 = 1.38$

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i1=εR[1−e−Rt/L] i2=dqdt=ddt[qo(1−e−t/RC)]i2=εRe−t/RC So, di2dt=−εR2Ce−t/RC , di1dt=εLe−Rt/L di1dt+di2dt=εLe−Rt/L−2εLe−2Rt/L=0 ⇒2eRto/L=e2Rto/L⇒ln2+RtoL=2RtoL ⇒to=2(RC)ln2 ⇒n=2ln2=1.38