Question

Question: Figure shows a battery connected across a uniform resistor R_-0. A sliding contact can move across the resistor from x = 0 at the left to x = 10 at the right. Moving the contact changes how much resistance is to the left of the contact and how much is to the right. Find the rate at which energy is dissipated in resistor as a function of x. Plot the function for $\mathit{\epsilon }\mathbf{=}\mathbf{50}\mathbf{\text{V}}\mathbf{,}\mathit{R}\mathbf{=}\mathbf{2000}\mathit{\Omega }\mathbf{,}$and${\mathit{R}}_{\mathbf{0}}\mathbf{=}\mathbf{100}\mathit{\Omega }\mathbf{,}$ .

Short answer

Answer:

The rate of the energy dissipated as a function of x is,
$P=\frac{100\times R{\left(\frac{x}{R_0}\right)}^2}{{\left(100\left(\frac{R}{R_0}\right)+10x-x^2\right)}^2}$

Step-by-step solution

Given data:

$$\begin{gathered} \text{Theresistor,}{R}_{}=2000\Omega \\ \text{Theresistor,}{R}_0=100\Omega \\ \text{Thevoltage,}\epsilon =50V \end{gathered}$$

Understanding the concept

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,

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

Find the rate at which energy is dissipated in resistor R  as a function of x : 

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.

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

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

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

Substitute 10 cm for x and L ,50 V for $\epsilon$. $2000\Omega$ for R , and $100\Omega$for $R_0$ in the above equation.

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

Step 4: Plot the function:

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