91Ó°ÊÓ

$$\begin{gathered} \text{Theresistor,}{R}_{}=2000\mathrm{Î\copyright } \\ \text{Theresistor,}{R}_0=100\mathrm{Î\copyright } \\ \text{Thevoltage,}\mathrm{Î\mu }=50V \end{gathered}$$

First find the equivalent resistance and then determine the voltage V across the resistor R. Once you know the equivalent resistance and voltage, define power P using the following formula.

Formulae:

The voltage is define by,

V = IR

The power is define by,

$P=\frac{V^2}{R}$

The resistance of the resistor R₀ connected in parallel with R . So,

$R=\frac{R_{0}x}{L}$

Now, define the equivalent resistance is as follow.

$$\begin{gathered} \frac{1}{R\text{'}}=\frac{1}{R}+\frac{1}{R_1} \\ =\frac{1}{R}+\frac{1}{\frac{R_{0}x}{L}} \\ R\text{'}=\frac{\frac{R{R}_{0}x}{L}}{R+\frac{R_{0}x}{L}} \end{gathered}$$

The equivalent resistance across R is,

$R_e=R_0\left(1-\frac{x}{L}\right)+R\text{'}$

Substitute $\frac{\frac{R{R}_{0}x}{L}}{R+\frac{R_{0}x}{L}}$ for R^' in the above equation.

$R_e=R_0\left(1-\frac{x}{L}\right)+\frac{\frac{R{R}_{0}x}{L}}{R+\frac{R_{0}x}{L}}$

The voltage across the resistor R is define as below.

$$\begin{gathered} V_R=\frac{\mathrm{Î\mu }zw3esR\text{'}}{R_e} \\ =\frac{\mathrm{Î\mu }\left(\frac{\frac{R{R}_{0}x}{L}}{R+\frac{R_{0}x}{L}}\right)}{R_0\left(1-\frac{x}{L}\right)+\left(\frac{\frac{R{R}_{0}x}{L}}{R+\frac{R_{0}x}{L}}\right)} \end{gathered}$$

Now, determine the power dissipated in resistor R as follow.

$$\begin{gathered} P=\frac{V_R^2}{R} \\ =\frac{1}{R}{\left(\frac{\mathrm{Î\mu }\left(\frac{\frac{R{R}_{0}x}{L}}{R+\frac{R_{0}x}{L}}\right)}{R_0\left(1-\frac{x}{L}\right)+\frac{\frac{R{R}_{0}x}{L}}{R+\frac{R_{0}x}{L}}}\right)}^2 \\ =\frac{1}{R}{\left(\frac{\mathrm{Î\mu }\left(\frac{R{R}_{0}x}{L}\right)}{R_0\left(1-\frac{x}{L}\right)+\left(R+\frac{R_{0}x}{L}\right)+\left(\frac{R{R}_{0}x}{L}\right)}\right)}^2 \\ =\frac{L^{2}R{\left(\frac{\mathrm{Î\mu }x}{R_0}\right)}^2}{{\left(L^2\left(\frac{R}{R_0}\right)+10x-x^2\right)}^2} \end{gathered}$$

Substitute 10 cm for x and L ,50 V for $\mathrm{Î\mu }$. $2000\mathrm{Î\copyright }$ for R , and $100\mathrm{Î\copyright }$for $R_0$ in the above equation.

$$\begin{gathered} P=\frac{{\left(10\text{cm}\right)}^2\left(2000\mathrm{Î\copyright }\right){\left(\frac{\left(50\text{V}\right)\left(10\text{cm}\right)}{100\mathrm{Î\copyright }}\right)}^2}{{\left({\left(10\text{cm}\right)}^2\left(\frac{2000\mathrm{Î\copyright }}{100\mathrm{Î\copyright }}\right)+10\left(10\text{cm}\right)-{\left(10\text{cm}\right)}^2\right)}^2} \\ =1.125\text{W} \end{gathered}$$

Here, you have assumed the values of x and from that the calculated power. And the graph of P vs X is plotted.