Question

Show that if$\mathit{L}\mathbf{,}\mathit{R}\mathbf{,}{\mathit{E}}_{\mathbf{\text{D}}}$, and$\mathit{\gamma }$are constant, then the amplitude of the steady-state current in Example 10 is a maximum when the capacitance is$\mathit{C}\mathbf{=}\mathbf{1}\mathbf{/}\mathit{L}{\mathit{\gamma }}^{\mathbf{2}}\mathbf{.}$

Short answer

We conclude that for the amplitude of the steady-state current to be maximum, we need $C=\frac{1}{L{\gamma }^2}$.

Step-by-step solution

Definition Of LRC Series circuits

LRC-SERIES CIRCUITS As was mentioned in the introduction to this chapter, many different physical systems can be described by a linear second-order differential equation similar to the differential equation of forced motion with damping:

$m\frac{d^{2}x}{d{t}^2}+\beta \frac{dx}{dt}+kx=f\left(t\right)$

Using LRC series

From example (10) in the book,

$i_p\left(t\right)=\frac{E_0}{Z}\left(\frac{R}{Z}\sin \gamma t-\frac{X}{Z}\cos \gamma t\right)$

where R is the circuit resistance,Z is the circuit impedance,X is the circuit reactance,$E_0$ is a constant related to the impressed voltage on the circuit and $\gamma$is the circuit frequency.

The circuit reactance is expressed as follows:

$X=L\gamma -\frac{1}{C\gamma }$

Circuit Impedance

Also, the circuit impedance is expressed as follows:

$Z=\sqrt{X^2+R^2}$

where L and C are the circuit inductance and capacitance respectively.

To maximize the amplitude of the steady-state current, we have to maximize$\frac{E_0}{Z}$as follows:

$\frac{E_0}{Z}=\frac{E_0}{\sqrt{X^2+R^2}}$

$\frac{E_0}{Z}=\frac{E_0}{\sqrt{X^2+R^2}}$

Since we need to maximize $\frac{E_0}{Z}$, we need to minimize Z, and since

$Z=\sqrt{R^2+{\left(L\gamma -\frac{1}{C\gamma }\right)}^2}$

Then, we need to minimize${\left(L\gamma -\frac{1}{C_{\gamma }}\right)}^2$ since R is just a constant. But${\left(L\gamma -\frac{1}{C_{\gamma }}\right)}^2$ is a positive definite, then the possible minimum value is 0 . Then,

$L\gamma -\frac{1}{C\gamma }=0$

We conclude that for the amplitude of the steady-state current to be maximum, we need .

$C=\frac{1}{L{\gamma }^2}$