# Find  i for t>0  if the circuit is in steady state at  t=0 ?

May 14, 2018

$i \left(t\right) = {i}_{\infty} \left(1 - {e}^{- \lambda t}\right)$

where ${i}_{\infty} = 6 \setminus \text{A}$ and $\lambda = 2 \setminus {\text{s}}^{-} 1$

#### Explanation:

At $t < 0$ the system is in steady state. So, the inductor behaves like a short and hence no current flows through the $4 \setminus \Omega$ resistance. So, the initial condition is

$i = 0$ at $t = 0$

For $t > 0$ the current through the $8 \setminus \Omega$ resistance is $9 \setminus \text{A} - i$, and so the voltage across it is

8\ Omega times (9\ "A"-i) = 72\ "V" - 8\ Omega times i

If the current through the inductance is ${i}^{'}$ then that through the $6 \setminus \Omega$ resistance is $i - {i}^{'}$ and so, we have

$72 \setminus \text{V" - 8\ Omega times i = 4\ Omega times i +6\Omega times (i-i^') + 6\ "V}$

and hence

$18 i - 6 {i}^{'} = 66 \setminus \text{A" implies i^' = 3i-11\ "A} \implies \frac{{\mathrm{di}}^{'}}{\mathrm{dt}} = 3 \frac{\mathrm{di}}{\mathrm{dt}}$

Again, by using the fact that the voltage drop across the inductor is given by $2 \setminus \text{H} \times \frac{{\mathrm{di}}^{'}}{\mathrm{dt}}$ we get

$72 \setminus \text{V" - 8\ Omega times i = 4\ Omega times i +2\ "H} \times \frac{{\mathrm{di}}^{'}}{\mathrm{dt}}$

So, the differential equation for $i \left(t\right)$ finally takes the form

$\frac{\mathrm{di}}{\mathrm{dt}} + \lambda i = \setminus \Lambda$

where $\setminus \lambda = 2 \setminus {\text{s}}^{-} 1$ and $\setminus \Lambda = 12 \setminus {\text{A"\ "s}}^{-} 1$

This differential equation can finally be cast in the form

$\frac{d}{\mathrm{dt}} \left(\frac{\Lambda}{\lambda} - i\right) = - \lambda \left(\frac{\Lambda}{\lambda} - i\right)$

which is easy to integrate to give

$\ln \left({i}_{\infty} - i\right) = - \lambda t + C$

where ${i}_{\infty} = \frac{\Lambda}{\lambda} = 6 \setminus \text{A}$. From the initial condition, we get

$C = \ln \left({i}_{\infty}\right)$

so that

$1 - \frac{i}{i} _ \infty = {e}^{- \lambda t}$ and thus finally

$i \left(t\right) = {i}_{\infty} \left(1 - {e}^{- \lambda t}\right)$