Question

(a) Show that the current i(t) in an L R C-series circuit satisfies

$\mathit{L}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{i}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathit{R}\frac{\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{C}}\mathit{i}\mathbf{=}{\mathit{E}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

where${\mathit{E}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$denotes the derivative of E(t).

(b) Two initial conditions i(0) and${\mathit{i}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$can be specified for the DE in part (a). If$\mathit{i}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{i}}_{\mathbf{0}}$and, $\mathit{q}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{q}}_{\mathbf{0}}$what is${\mathit{i}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$?

Short answer

(a)Use Kirchhoff's second law and the fact that$i=\frac{dq}{dt}$to verify the equation.

(b) Substitute initial conditions in the equation to get$i^{\text{'}}\left(0\right)=\frac{1}{L}\left[E\left(0\right)-R{i}_0-\frac{q_0}{C}\right]$

Step-by-step solution

Step 1:Definition of Non-linear and Linear Spring

NONLINEAR SPRINGS The mathematical model has the form

$m\frac{d^{2}x}{d{t}^2}+F\left(x\right)=0$

where$F\left(x\right)=kx$. Becausedenotes the displacement of the mass from its equilibrium position,$F\left(x\right)=kx$is Hooke's law-that is, the force exerted by the spring that tends to restore the mass to the equilibrium position. A spring acting under a linear restoring force$F\left(x\right)=kx$is naturally referred to as a linear spring.

A spring whose mathematical model incorporates a nonlinear restorative force, such as

$m\frac{d^{2}x}{d{t}^2}+k{x}^3=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}m\frac{d^{2}x}{d{t}^2}+kx+k_1{x}^3=0$

is called a nonlinear spring.

Using the LRC series

(a) In a$LRC$-series electrical circuit, by Kirchhoff's second law the sum of voltage drops across the inductor, resistor, and capacitor equals the voltage$E\left(t\right)$impressed on the circuit, that is

$L\frac{di}{dt}+Ri+\frac{1}{C}q=E\left(t\right)$

Upon taking the derivative with respect to time, we obtain

$L\frac{d^{2}i}{d{t}^2}+R\frac{di}{dt}+\frac{1}{C}\frac{dq}{dt}=E^{\text{'}}\left(t\right)$

Since chargeon the capacitor is related to the current$i\left(t\right)$by$i=\frac{dq}{dt}$, so the above equation becomes

$L\frac{d^{2}i}{d{t}^2}+R\frac{di}{dt}+\frac{1}{C}i=E^{\text{'}}\left(t\right)$

Substitute the initial conditions

(b) Substituting the initial conditions$i\left(0\right)=i_0$and $q\left(0\right)=q_0$in equation (1) now gives

$$\begin{gathered} L{i}^{\text{'}}\left(0\right)+Ri\left(0\right)+\frac{1}{C}q\left(0\right)=E\left(0\right) \\ L{i}^{\text{'}}\left(0\right)+R{i}_0+\frac{1}{C}{q}_0=E\left(0\right) \\ L{i}^{\text{'}}\left(0\right)=E\left(0\right)-R{i}_0-\frac{q_0}{C} \end{gathered}$$

$i^{\text{'}}\left(0\right)=\frac{1}{L}\left[E\left(0\right)-R{i}_0-\frac{q_0}{C}\right]$