Question

In Fig. 27-53, the resistors have the values ${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\Omega }$, ${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\Omega }$, and ${\mathbf{R}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\Omega }$, and the ideal battery’s emf is$\mathbf{\epsilon }\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$. For what value of R₄will the rate at which the battery transfers energy to the resistors equal (a)60.0 W, (b) the maximum possible rate P_max, and (c) the minimum possible rate P_min? What are (d)P_maxand (e)P_min?

Short answer

1. The value of at which the rate of energy transferred to the resistors will be 60 W is $19.5\mathrm{\Omega }$.
2. The resistance value for maximum possible rate P_max is 0.
3. The resistance value for minimum possible rate P_min is $\infty$.
4. The value of maximum possible rate P_max is 82.3 W.
5. The value of minimum possible rate P_min is 57.6 W.

Step-by-step solution

The given data

The value of the resistors,${\mathrm{R}}_1=7\mathrm{\Omega }$,${\mathrm{R}}_2=12\mathrm{\Omega }$,${\mathrm{R}}_3=4\mathrm{\Omega }$, and${\mathrm{R}}_4=\mathrm{R}\mathrm{\Omega }$

Ideal battery emf,$\mathrm{\epsilon }=24.0\mathrm{V}$

Rate at which the battery transfers energy to the resistors equally,$\mathrm{P}=60.0\mathrm{W}$

Understanding the concept of energy rate

The rate of energy is given by the ratio of the square of the emf to the equivalent resistance. Using the equivalence resistance concept for the given combination of the resistors, the equation of power to the resistance can be given. We can now get the resistance value for each given case and using the resistance value, the power of the combination can be given.

Formulae:

The rate of energy transfer or the power output, $\mathrm{P}=\frac{{\mathrm{\epsilon }}^2}{\mathrm{R}}$ (1)

The equivalent resistance for series combination of the resistors, data-custom-editor="chemistry" ${\mathrm{R}}_{\mathrm{eq}}=\underset{\mathrm{i}}{\overset{\mathrm{n}}{\sum {\mathrm{R}}_{\mathrm{i}}}}$ (2)

The equivalent resistance for parallel combination of the resistors, data-custom-editor="chemistry" ${\mathrm{R}}_{\mathrm{eq}}=\underset{\mathrm{i}}{\overset{\mathrm{n}}{\sum \frac{1}{{\mathrm{R}}_{\mathrm{i}}}}}$ (3)

a) Calculation of the resistance R4

The equivalent resistance of the given combination of resistors can be given using equations (2) and (3) as follows:

(R₁ is in series combination with the resistors R₂,R₃ and R₄ in parallel)

$\begin{array}{rcl}{\mathrm{R}}_{\mathrm{eq}}& =& {\mathrm{R}}_1+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}+\frac{1}{{\mathrm{R}}_4}\\ & =& {\mathrm{R}}_1+\frac{{\mathrm{R}}_2{\mathrm{R}}_3{\mathrm{R}}_4}{{\mathrm{R}}_3{\mathrm{R}}_2+{\mathrm{R}}_3{\mathrm{R}}_4+{\mathrm{R}}_2{\mathrm{R}}_4}\\ & =& 7.00\mathrm{\Omega }+\frac{\left(12\mathrm{\Omega }\right)\left(4\mathrm{\Omega }\right)\mathrm{R}}{\left(4\mathrm{\Omega }\right)\left(12\mathrm{\Omega }\right)+\left(4\mathrm{\Omega }\right)\mathrm{R}+\left(12\mathrm{\Omega }\right)\mathrm{R}}\\ & =& 7.00\mathrm{\Omega }+\frac{\left(48{\mathrm{\Omega }}^2\right)\mathrm{R}}{\left(48{\mathrm{\Omega }}^2\right)+\mathrm{R}\left(16\mathrm{\Omega }\right)}.........................\left(4\right)\end{array}$

Now, using the above equivalent resistance value in equation (1), the value of resistance can be given as follows:

$\begin{array}{rcl}60\mathrm{W}& =& \frac{{\left(24\mathrm{V}\right)}^2}{7.00\mathrm{\Omega }+{\displaystyle \frac{\left(48{\mathrm{\Omega }}^2\right)\mathrm{R}}{\left(48{\mathrm{\Omega }}^2\right)+\mathrm{R}\left(16\mathrm{\Omega }\right)}}}\\ 60\mathrm{W}\left(7.00\mathrm{\Omega }+\frac{\left(3\mathrm{\Omega }\right)\mathrm{R}}{\left(3\mathrm{\Omega }\right)+\mathrm{R}}\right)& =& {\left(24\mathrm{V}\right)}^2\\ 5\left(7.00\mathrm{\Omega }+\frac{\left(3\mathrm{\Omega }\right)\mathrm{R}}{\left(3\mathrm{\Omega }\right)+\mathrm{R}}\right)& =& 48\frac{{\mathrm{V}}^2}{\mathrm{W}}\\ \left(35.00\mathrm{\Omega }+\frac{\left(15\mathrm{\Omega }\right)\mathrm{R}}{\left(3\mathrm{\Omega }\right)+\mathrm{R}}\right)& =& 48\mathrm{\Omega }\\ \frac{\left(15\mathrm{\Omega }\right)\mathrm{R}}{\left(3\mathrm{\Omega }\right)+\mathrm{R}}& =& 13\mathrm{\Omega }\\ \left(15\mathrm{\Omega }\right)\mathrm{R}& =& \left(13\mathrm{\Omega }\right)\mathrm{R}+39{\mathrm{\Omega }}^2\\ \left(2\mathrm{\Omega }\right)\mathrm{R}& =& 39{\mathrm{\Omega }}^2\\ \mathrm{R}& =& \frac{39{\mathrm{\Omega }}^2}{2\mathrm{\Omega }}\\ \mathrm{R}& =& 19.5\mathrm{\Omega }\end{array}$

Hence, the value of the fourth resistance is data-custom-editor="chemistry" $19.5\mathrm{\Omega }$.

b) Calculation of the resistance for maximum possible rate

From equation (1), we get that.$\mathrm{P}\propto \frac{1}{\mathrm{R}}$

Thus, we can get the maximum rate value atdata-custom-editor="chemistry" $\mathrm{R}=0$ as follows:data-custom-editor="chemistry" $\mathrm{P}=\infty$

Hence, the resistance value for the maximum possible rate is 0.

c) Calculation of the resistance for minimum possible rate

From equation (1), we get that $\mathrm{P}\propto \frac{1}{\mathrm{R}}$.

Thus, we can get the maximum rate value at$\mathrm{R}=\infty$ as follows:$\mathrm{P}=0$

Hence, the resistance value for the maximum possible rate is $\infty$.

d) Calculation of the maximum possible rate

The maximum possible value can be given equation (1) for${\mathrm{R}}_{\mathrm{eq},\min }=7.00\mathrm{\Omega }$ as follows:

$\begin{array}{rcl}{\mathrm{P}}_{\max }& =& \frac{{\left(24\mathrm{V}\right)}^2}{7.00\mathrm{\Omega }}\\ & =& 82.3\mathrm{W}\end{array}$

Hence, the value of maximum rate is 82.3 W.

e) Calculation of the minimum possible rate

The equivalent resistance can be given using equation (2) and (3) for R=0 as follows:

$\begin{array}{rcl}{\mathrm{R}}_{\mathrm{eq},\max }& =& 7.00\mathrm{\Omega }+\frac{\left(12\mathrm{\Omega }\right)\left(4\mathrm{\Omega }\right)}{\left(12\mathrm{\Omega }\right)+\left(4\mathrm{\Omega }\right)}\\ & =& 7.00\mathrm{\Omega }+3.00\mathrm{\Omega }\\ & =& 10.00\mathrm{\Omega }\end{array}$

The minimum possible value can be given equation (1) fordata-custom-editor="chemistry" ${\mathrm{R}}_{\mathrm{eq},\max }=10\mathrm{\Omega }$ as follows:

$\begin{array}{rcl}{\mathrm{P}}_{\min }& =& \frac{{\left(24\mathrm{V}\right)}^2}{10.0\mathrm{\Omega }}\\ & =& 57.6\mathrm{W}\end{array}$

Hence, the value of maximum rate is 57.6 W.