Question

A series circuit contains an inductance of,$\mathit{L}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{\text{h}}$ a capacitance of, $\mathit{C}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\text{f}}$and an electromotive forceof$\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathit{s}\mathit{i}\mathit{n}\mathbf{50}\mathit{t}\mathbf{}\mathbf{\text{V}}$. Initially, the charge q and current i are zero.

(a) Determine the charge q(t).

(b) Determine the current i(t).

(c) Find the times for which the charge on the capacitor is zero.

Short answer

(a) \(q(t)=\dfrac{2}{15}\sin(50t)\) (b) \(i(t)=\dfrac{20}{3}\cos(50t)\) (c) \(t=\dfrac{k\pi}{100},\, k\in \mathbb Z\).

Step-by-step solution

Definition of Non-linear and Linear Spring 

NONLINEAR SPRINGS

The mathematical model has the form

where. Becausedenotes the displacement of the mass from its equilibrium position,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 forceis naturally referred to as a linear spring.

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

is called a nonlinear spring.

Find the auxiliary equation

We were given the next informations:

-$L=1\text{h}$

-$C={10}^{-4}\text{f}$

-$q\left(0\right)=0$

-$i\left(0\right)=q^{\text{'}}\left(0\right)=0$

- The electromotive force$E\left(t\right)=100\sin 50t\text{V}$is impressed on the system.

The equation that represents the circuit is:

$q^{\text{'}\text{'}}+{10}^{4}q=100\sin 50t$

The zeros of the auxiliary equation are

$x_{1,2}=\pm 100i$

Therefore, the homogeneous general solution of Eq. (1) is

$q_h\left(t\right)=c_1\cos 100t+c_2\sin 100t$

Substitute the initial condition

From the first initial condition, we get.$c_1=0$ Let us derive (2):

$q_h^{\text{'}}\left(t\right)=-100c_1\sin 100t+100c_2\cos 100t$

From the second initial condition, we get

$c_2=0$

The homogeneous solution is

$x_h\left(t\right)=0$

Let us assume that the particular solution is of the form

$q_p\left(t\right)=A\sin 50t+B\cos 50t$

Its derivations are

$q_p^{\text{'}}\left(t\right)=50(A\cos 50t-B\sin 50t),q_p^{\text{'}\text{'}}\left(t\right)=250(-A\sin 50t-B\sin 50t)$

Substitute it into (1):

$250(-A\sin 50t-B\sin 50t)+1000(A\sin 50t+B\cos 50t)=100\sin 50t\iff$

$750(A\sin 50t+B\cos 50t)=100\sin 50t$

The particular solution

From this, we get that$B=0$and.$A=\frac{10}{75}$Therefore, the particular solution is

$q_p\left(t\right)=\frac{10}{75}\sin 50t$

The solution is given by$q\left(t\right)=q_h\left(t\right)+q_p\left(t\right)$:

$q\left(t\right)=\frac{2}{15}\sin 50t$

(b)

Since,$i\left(t\right)=q^{\text{'}}\left(t\right)$ simply derive (5) to find$i\left(t\right):q^{\text{'}}\left(t\right)=i\left(t\right)=\frac{20}{3}\cos 50t$

(c)

The charge is zero when $\sin 50t$is zero, that is when

$50t=\frac{k\pi }{2},k\in \mathbb{Z},\iff t=\frac{k\pi }{100},k\in \mathbb{Z}$