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202206081501 Exercise 4.1 (Q15)

In an electric circuit with a supplied voltage (emf) E, a resistor with resistance r_0, and an inductor with reactance x_0, suppose you want to add a second resistor. If r represents the resistance of this second resistor then the power P delivered to that resistor is given by

\displaystyle{P=\frac{E^2r}{(r+r_0)^2+x_0^2}}

with E, r_0, and x_0 treated as constants. For which value of r is the power P maximized?

extracted from Michael Corral. (2020). Elementary Calculus

Lemma. (quotient rule)

Let f(x)=\displaystyle{\frac{g(x)}{h(x)}}, where both g and h are differentiable and h(x)\neq 0. The quotient rule states that the derivative of f(x) is

f'(x)=\displaystyle{\frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}}

Wikipedia on Quotient Rule

Roughwork.

Defining

\begin{aligned} g(r) & = E^2r\\ h(r) & = (r+r_0)^2+x_0^2\\ \end{aligned}

and

\begin{aligned} g'(r) & = E^2\\ h'(r) & = 2(r+r_0)\\ \end{aligned}

then

\displaystyle{\frac{\mathrm{d}P}{\mathrm{d}r}}=0\Longrightarrow g'(r)h(r)-g(r)h'(r)=0

Computing as follows

\begin{aligned} (E^2)\big( (r+r_0)^2+x_0^2 \big) & = (E^2r)\big( 2(r+r_0) \big) \\ (r+r_0)^2+x_0^2 & = 2r(r+r_0) \\ r^2+2rr_0+r_0^2 +x_0^2 & = 2r^2+2rr_0 \\ r & = \sqrt{r_0^2+x_0^2} \\ \end{aligned}

The mathematically formal way is to show that

\displaystyle{\frac{\mathrm{d}^2P(r)}{\mathrm{d}r}\bigg|_{r=\sqrt{r^2_0+x_0^2}}}<0

But from a physical point of view, assume that the electric currents i passing through every components in series are the same, and the potential difference across each total up to the supplied voltage, namely,

\begin{aligned} i(t) & = I_{\textrm{peak}}\sin (\omega t) \textrm{ where }I_{\textrm{peak}}\textrm{= const.} \\ E & = ir_0+x_0\frac{\mathrm{d}i}{\mathrm{d}t}+ir \\ \end{aligned}

The power P delivered to resistor r is given by the formula

\boxed{P=Vi=\displaystyle{\frac{V^2}{r}}=i^2r}

and the only way to maximize P, is to maximize either or both i and V.

Solving for a first-order ordinary differential equation:

f(r,i,i')= ir_0+x_0i'+ir-E=0

deriving current i wrt \omega t, we have

\begin{aligned} 0 & = i'r_0+x_0i''+i'r \\ 0 & = -x_0I_{\textrm{peak}}\sin (\omega t)+(r_0+r)I_{\textrm{peak}}\cos (\omega t) \\ 0 & = -x_0\tan (\omega t) + r_0+r \\ \tan (\omega t) & = \frac{r_0+r}{x_0} \\ \sin (\omega t)&=\frac{r_0+r}{\sqrt{(r_0+r)^2+x_0^2}} \\ \end{aligned}

Thus i(r)=\displaystyle{I_{\textrm{peak}}\frac{r_0+r}{\sqrt{(r_0+r)^2+x_0^2}}}

\begin{aligned} P & = i^2r \\ & = \bigg(\frac{I_{\textrm{peak}}(r_0+r)}{\sqrt{(r_0+r)^2+x_0^2}}\bigg)^2 \cdot r \\ & = \frac{I_{\textrm{peak}}^2(r_0+r)^2r}{(r_0+r)^2+x_0^2} \\ \end{aligned}

(to be continued)

Recall the relation between root mean square (rms) values and peak values:

\begin{aligned} V_{\textrm{rms}} & = \frac{V_\textrm{peak}}{\sqrt{2}} \\ I_{\textrm{rms}} & = \frac{I_\textrm{peak}}{\sqrt{2}} \\ \end{aligned}

Recall also that the resistance R_L of an ideal inductor is zero (=0), and that after the circuit has shortly reached steady state (i.e., constant current i anywhere/anytime), the potential difference (\textrm{p.d.}) or voltage drop (\Delta V) across the inductor will become zero (=0) before long.

Try again,

\begin{aligned} I_{\textrm{peak}} & = i \\ \frac{I_{\textrm{peak}}^2(r_0+r)^2r}{(r_0+r)^2+x_0^2} & = \frac{i^2(r_0+r)^2r}{(r_0+r)^2+x_0^2} \\ & = \frac{(ir_0+ir)^2r}{(r_0+r)^2+x_0^2} \\ & = \frac{E^2r}{(r_0+r)^2+x_0^2} \\ \end{aligned}

The derivative test seems inevitable. Maybe you could show that resistance r=\sqrt{r_0^2+x_0^2} maximizes power P, simply by drawing a phasor diagram?

(to be continued)