# Transient analysis of first order circuit

For the attached problem, I am getting the signs wrong.Here I have considered the inductor voltage signs as such because after closing the switch, less current would flow through the inductor but because of Lenz's law , it would be try to resist that.Where am I getting this wrong? Any help would be appreciated. Thanks

EDIT- This is problem 7.54 of Fundamentals of Electric Circuits by Charles Alexander and Matthew Sadiku 7th edition. The correct answer that I have given here is from the book itself.

• Initial conditon at t = 0 for inductor current is 1A so it seems like both answers are wrong. Commented Jul 13 at 8:05
• I made a mistake substitiung t= 0 to check the answers but you deleted your comment. Commented Jul 13 at 12:39

Just before $$\t=0\$$ you know that the initial condition for the inductor is as follows:

simulate this circuit – Schematic created using CircuitLab

Once the switch is flipped you have:

simulate this circuit

The above can be replaced with:

simulate this circuit

It follows from KCL that:

\begin{align*} \frac{v_t}{R_\text{TH}}+\frac{v_t}{R_2}&= \frac{v_L}{R_2}+I_1 \end{align*}

Or, that:

\begin{align*} v_t&= v_L\frac{R_\text{TH}}{R_\text{TH}+R_2}+I_1\cdot\left(R_\text{TH}\vert\vert R_2\right) \end{align*}

You also know that:

\begin{align*} \frac{v_L}{R_2}+\frac1{L_1}\int v_L\:\text{d}t&=\frac{v_t}{R_2} \\\\ v_t&=v_L+\frac{R_2}{L_1}\int v_L\:\text{d}t \end{align*}

Setting these equal:

\begin{align*} v_L\frac{R_\text{TH}}{R_\text{TH}+R_2}+I_1\cdot\left(R_\text{TH}\vert\vert R_2\right)&=v_L+\frac{R_2}{L_1}\int v_L\:\text{d}t \\\\ \frac{R_\text{TH}}{R_\text{TH}+R_2}\frac{\text{d}}{\text{d}t}v_L&=\frac{\text{d}}{\text{d}t}v_L+\frac{R_2}{L_1}v_L \\\\ \left[\frac{\text{d}}{\text{d}t}+\frac{R_\text{TH}+R_2}{L_1}\right]v_L&=0 \end{align*}

The solution to this is:

$$v_L=A_1\exp\left(-\frac{R_\text{TH}+R_2}{L_1} t\right)$$

With $$\A_1=-1\$$, obviously. So $$\v_L=-\exp\left(-2 t\right)\$$.

You know that $$\i_L=\frac1{L_1}\int v_L\:\text{d}t=\frac17\exp\left(-2 t\right)+A_2\$$. Obviously, $$\A_2=\frac67\$$. So:

\begin{align*} i_L&=\frac17\exp\left(-2 t\right)+\frac67 \\\\ &=\frac17\left[6+\exp\left(-2 t\right)\right] \end{align*}

The answer, $$\i_L=\frac17\left[6-\exp\left(-2 t\right)\right]\$$, is wrong.

• Thanks for the detailed answer. Just one question, if I change the direction of inductor voltage in my solution, I get the same answer as yours. Does that mean that in these type of problems, I should keep the higher voltage at that inductor end where the current is entering ? Commented Jul 13 at 11:43
• @lefty The answer is always the same as far as the current direction and magnitude goes. However, if you applied a voltmeter across it as shown it would read the negation of what I wrote. If you inserted an ammeter in series but oriented that way you would get the negation, too. I may have misread what the author wanted. The physics and behavior is the same. It may just be the author wanted meters wired opposite of normal. If so, just negate the two results I gave. A1 and A2 would have opposite signs, is all. But it isn't conventional. Maybe the author intended that, though. Commented Jul 13 at 14:38

There are two forms for 1st order transients that are very useful. They can be used for checking your work, or understanding the problem or even solving the problem sometimes by inspection. The first form separates the transient solution from the steady state solution. $$y(t)=\left(Y_{I}-Y_{F}\right)e^{\frac{-t}{\tau}}+Y_{F}$$ and the second one identifies discharging the initial value while charging the final value. $$y(t)=Y_{I}e^{\frac{-t}{\tau}}+Y_{F}\left(1-e^{\frac{-t}{\tau}}\right)$$

For the circuit in the OP, $$\y(t)=i_{L}(t)\$$, $$\Y_{I}=I_{LI}=1\textrm{A}\$$ is the initial value, and $$\Y_{F}=I_{LF}=\frac{6}{7}\textrm{A}\$$ is the final value. The time constant is $$\\tau=\frac{L}{R_{N}}=\frac{3.5}{7}=\frac{1}{2}\$$. $$\R_{N}\$$ is the Norton resistance seen by the inductance.

Using the first equation: $$i_{L}(t)=\left(1-\frac{6}{7}\right)e^{-2t}+\frac{6}{7}$$ $$i_{L}(t)=\frac{1}{7}\left(6+e^{-2t}\right)$$