Question

Question: In Figure, the resistances are,${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\Omega }$,${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\Omega }$ and the battery is ideal. What value of R₃maximizes the dissipation rate in resistance 3?

Short answer

Answer

$1.43\Omega$is the value for R₃ , which maximizes the dissipation rate inR₃

Step-by-step solution

Given

$$\begin{gathered} R_1=2.00\Omega \\ {R}_2=5.00\Omega \end{gathered}$$

Understanding the concept

First, we have to find the power across the resistor.For maximum dissipation, the power must be maximum. So, we have to differentiate the equation of power respective toand equate this to zero. From this, we can find the value for.

Formula:

$$\begin{gathered} P=\frac{V^2}{R} \\ V=IR \end{gathered}$$

Calculate the value of  that maximizes the dissipation rate in resistance

We calculate the equivalent resistance for R₃ and R₂,

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_3}+\frac{1}{R_2} \\ \frac{1}{R_{eq}}=\frac{1}{R_3}+\frac{1}{5} \\ {R}_{eq}=\frac{1}{\frac{1}{R_3}+\frac{1}{5}} \\ {R}_{eq}=\frac{5R_3}{5+R_3} \end{gathered}$$

Now,

is in parallel with the

So,

$$\begin{gathered} R_s=R_1+R_{eq} \\ {R}_s=2+\left(\frac{5R_3}{5+R_3}\right) \\ {R}_s=\frac{10+7R_3}{5+R_3} \end{gathered}$$

Current flowing through R_S will be

$$\begin{gathered} I=\frac{\epsilon }{R} \\ I=\frac{\epsilon \left(5+R_3\right)}{\left(10+7R_3\right)} \end{gathered}$$

The voltage across whole circuit will be

$$\begin{gathered} V=IR \\ {V}_s=\left(\frac{\epsilon \left(5+R_3\right)}{\left(10+7R_3\right)}\right)\times \left(\frac{5R_3}{5+R_3}\right) \\ {V}_s=\frac{\epsilon \left(5R_3\right)}{10+7R_3} \end{gathered}$$

So, the power across R₃ can be given as

$$\begin{gathered} P=\frac{V_s^2}{R_3} \\ P=\frac{{\left(\frac{\epsilon \left(5R_3\right)}{10+7R_3}\right)}^2}{R_3} \\ P=\frac{{\epsilon }^2}{R_3}\times \frac{25R_3^2}{{\left(10+7R_3\right)}^2} \end{gathered}$$

For the maximum dissipation, the power must be maximum. Therefore,

$$\begin{gathered} \frac{dP}{d{R}_3}=0 \\ \frac{d\left(\frac{{\epsilon }^2}{R_3}\times \frac{25R_3^2}{{\left(10+7R_3\right)}^2}\right)}{d{R}_3}=0 \\ 25{\epsilon }^2\times \left(\frac{{\left(10+7R_3\right)}^2-\left(2R_3\left(10+7R_3\right)\left(7\right)\right)}{{\left(10+7R_3\right)}^4}\right)=0 \\ \left(\frac{{\left(10+7R_3\right)}^2-\left(14R_3\left(10+7R_3\right)\right)}{{\left(10+7R_3\right)}^4}\right)=0 \\ {\left(10+7R_3\right)}^2-\left(14R_3\left(10+7R_3\right)\right)=0 \\ 10+7R_3-14R_3=10-7R_3=0 \\ {R}_3=\frac{10}{7} \\ {R}_3=1.43\Omega \end{gathered}$$

As we know,

$P\propto \frac{1}{R_3}$

So, the lower value R₃ for will result into higher value for P

Thus, for the maximum value of P , we consider the lowest value for R₃ , which is $R_3=1.43\Omega$.