Question

The switch in FIGURE P30.77 has been open for a long time. It is closed at $t=0\mathrm{s}$.

a. After the switch has been closed for a long time, what is the current in the circuit? Call this current $I_0$.

b. Find an expression for the current $I$as a function of time. Write your expression in terms of $I_{0,},R,\text{and}L$.

c. Sketch a current-versus-time graph from $t=0\mathrm{s}$until the current is no longer changing.

Short answer

(a) The current in the circuit$I_o=\frac{\mathrm{\Delta }{V}_{\mathrm{bat}}}{R}$

(b) An expression for the current$I=I_o\left(1-e^{\frac{-R}{L}t}\right)$

(c) Draw an exponential curve for the increase of the current till becomes constant at$I_o$

Step-by-step solution

Inductor

(a)As time passes, the current that flows through the inductance changes. The current through the inductor remains zero more than a long period of time as it becomes stable. As a response, the potential difference throughout the inductor is zero at about this position.

$\mathrm{\Delta }{V}_L=-L\left(\frac{dI}{dt}\right)=-L(0\mathrm{A}/\mathrm{s})=0$

The inductor functions as a short-circuit wire whenever the current across it is zero. The current flowing in this instance seems to be the flow related to the resistance.

we use Ohm's Law,

$I_o=\frac{\mathrm{\Delta }{V}_{\mathrm{bat}}}{R}$

Resistor and Inductor

(b) The current passing into both the resistors and the inductor if it flows through the circuit. Through into the inductor, the current changes. The loop rule, which states that now the summed electric potential all over the loop is zero, can be determined from Kirchhoff's law.

$\mathrm{\Delta }{V}_{\text{bat}}-IR-L\frac{dI}{dt}=0$

The value of current passing through into the batteries was $I_o$. So, the potential difference across the battery.

$\begin{array}{c}{I}_{o}R-IR-L\frac{dI}{dt}=0\\ R\left(I_o-I\right)=L\frac{dI}{dt}\\ \frac{dI}{\left(I_o-I\right)}=\frac{R}{L}dt\end{array}$

So we integrate,

$\begin{array}{c}{\int }_0^{I_o} \frac{dI}{\left(I_o-I\right)}=\frac{R}{L}{\int }_0^t dt\\ -{\left[\ln \left(I_o-I\right)\right]}_0^{I_o}=\frac{R}{L}t\\ \ln \left(\frac{I_o-I}{I_o}\right)=\frac{-R}{L}t\\ \left(\frac{I_o-I}{I_o}\right)=e^{\frac{-R}{L}t}\\ I=I_o\left(1-e^{\frac{-R}{L}t}\right)\end{array}$

 Exponential Curve

(c) Whenever the device is turned off, the current flowing starts at zero and gets larger per the part of the existing equation (a). The current increases till the reach a stable for which the inductor current matches the circuit's maximum current $I=I_o$. SAs a result, we will create an exponential curve for the growth of current till the reaches the point where this is constant.