Question

Find the solutions of (1.2)$\left(putI=dqIdt\right)$and (1.3), if $\mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{0}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\omega }\mathbf{\text{'}}\mathit{t}$( const.).

Short answer

Answer

The solution is $L\frac{d^{2}l}{d{t}^2}+R\frac{dl}{dt}+\frac{I}{C}-V_0\omega \text{'}\cos \omega \text{'}t$.

Step-by-step solution

Given information from question

Given values are $I=dqIdt)$and$V=V_u\sin \omega \text{'}t$

The differential equation satisfied by the current in a simple series circuit

The differential equation satisfied by the current in a simple series circuit is:

$\mathit{L}\frac{\mathbf{d}\mathbf{l}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{R}\mathit{I}\mathbf{+}\frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathit{V}$

The wide range of topics giving to differential equation

Given that,

$V=V_0\sin \omega \text{'}t\left(\omega \text{'}=cons\tan t\right)$ ……… (1)

The differential equation satisfied by the current in a simple series circuit is:

From equation (1)

$L\frac{dt}{dt}+RI+\frac{q}{C}=V_0\sin \omega \text{'}t$

Differentiating this equation with respect to t and substitute $\frac{dq}{dt}=I$

$$\begin{gathered} L\frac{d^{2}I}{d{t}^2}+R\frac{dI}{dt}+\frac{I}{C}=\frac{I}{C}\left(V_0\sin \omega \text{'}t\right) \\ L\frac{d^{2}I}{d{t}^2}+R\frac{dI}{dt}+\frac{I}{C}=V_0\omega \text{'}\cos \omega \text{'}t \end{gathered}$$

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Thus, the solution is $L\frac{d^{2}I}{d{t}^2}+R\frac{dt}{dt}+\frac{I}{C}=V_0\omega \text{'}t$.