In the circuit shown in Fig. E26.20, the rate at which R₁ is dissipating electrical energy is 15.0 W. (a) Find R₁ and R₂. (b) What is the emf of the battery? (c) Find the current through both R₂ and the 10.0 Ω resistor. (d) Calculate the total electrical power consumption in all the resistors and the electrical power delivered by the battery. Show that your results are consistent with conservation of energy.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. Introducing the constant of proportionality, the resistance one arrives at the usual mathematical equation that describes this relationship

Given Data

The current is 2.0 A in resistance R₁ and the current of 3.5 A is the total current flow in the circuit.

Dissipative power is = 15.0 W

(a)

The dissipative power can be calculated using the expression,

$$\begin{gathered} {\mathrm{R}}_1=\frac{{\mathrm{P}}_1}{{\mathrm{I}}_1^2} \\ {\mathrm{R}}_1=\frac{15.0\mathrm{W}}{{(2.0\mathrm{A})}^2} \\ {\mathrm{R}}_1=3.75\mathrm{\Omega } \end{gathered}$$

The three resistors R₁, R₂, and${\mathrm{R}}_3=10.0\mathrm{\Omega }$ are in parallel, and the potential difference V across resistors connected in the parallel is the same for every resistor and equals the potential difference across the combination, therefore,

${\mathrm{V}}_1={\mathrm{V}}_2={\mathrm{V}}_3=\mathrm{\epsilon }$

Where V1 could be calculated by Ohm's law by using the values for I₁ and R₁ such that,

$$\begin{gathered} {\mathrm{V}}_1={\mathrm{I}}_1{\mathrm{R}}_1 \\ {\mathrm{V}}_1=2\mathrm{A}\times 3.75\mathrm{\Omega } \\ {\mathrm{V}}_1=7.5\mathrm{V} \end{gathered}$$

Now We can use this voltage to get current flowing in the resistance ${\mathrm{R}}_3=10.0\mathrm{\Omega }$

$$\begin{gathered} {\mathrm{I}}_3=\frac{{\mathrm{V}}_3}{10.0\mathrm{\Omega }} \\ {\mathrm{I}}_3=\frac{7.5\mathrm{V}}{10.0\mathrm{\Omega }} \\ {\mathrm{I}}_3=0.75\mathrm{A} \end{gathered}$$

The total current 3.5 A is cut into the three resistors,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{t}}={\mathrm{l}}_1+{\mathrm{l}}_2+{\mathrm{l}}_3 \\ {\mathrm{l}}_2=3.5\mathrm{A}-2\mathrm{A}-0.75\mathrm{A} \\ {\mathrm{I}}_2=0.75\mathrm{A} \end{gathered}$$

Now We can get R₂ using Ohm's law such that,

$$\begin{gathered} {\mathrm{R}}_2=\frac{{\mathrm{V}}_2}{{\mathrm{I}}_2} \\ =\frac{7.5\mathrm{V}}{0.75\mathrm{A}} \\ =10\mathrm{\Omega } \end{gathered}$$

Therefore the resistance is ${\mathrm{R}}_1=3.75\mathrm{\Omega }\mathrm{and}{\mathrm{R}}_2=10.0\mathrm{\Omega }$

(b)

We want to find the emf of the battery.

As we discussed in part (a) the current through resistors connected in parallel is

the sum of each resistor, so, we can use the value of It to get the emf where the emf of the battery in the case of the equivalent resistance in the circuit R₁, and R₂ are in parallel so

We can use the following expression,

Then we calculate the equivalent resistance, such that,

$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{equivalent}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3} \\ \frac{1}{{\mathrm{R}}_{\mathrm{equivalent}}}=\frac{1}{3.75\mathrm{\Omega }}+\frac{1}{10\mathrm{\Omega }}+\frac{1}{10\mathrm{\Omega }} \\ \frac{1}{{\mathrm{R}}_{\mathrm{equivalent}}}=\frac{7}{15\mathrm{\Omega }} \end{gathered}$$

Using the above expression, we can write,

$$\begin{gathered} \mathrm{\epsilon }={\mathrm{I}}_{\mathrm{t}}{\mathrm{R}}_{\mathrm{equivalent}} \\ \mathrm{\epsilon }=3.5\hspace{0.33em}\mathrm{A}\times 2.14\hspace{0.33em}\mathrm{\Omega } \\ \mathrm{\epsilon }=7.5\hspace{0.33em}\mathrm{V} \end{gathered}$$

(c)

Above we have calculated the current through R₂ and R₃, such that

${\mathrm{I}}_2={\mathrm{I}}_3=0.75\hspace{0.33em}\mathrm{A}$

(d)

We can get the total power dissipated in the three resistors such that,$$\begin{gathered} \mathrm{P}={\mathrm{I}}^2\mathrm{R} \\ {\mathrm{P}}_{\mathrm{t}}={\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3 \\ {\mathrm{P}}_{\mathrm{t}}=15\hspace{0.33em}\mathrm{W}+\left[\right(0.75\hspace{0.33em}\mathrm{A}{)}^{2}10\hspace{0.33em}\mathrm{\Omega }]+[(0.75\hspace{0.33em}\mathrm{A}{)}^{2}10\hspace{0.33em}\mathrm{\Omega }] \\ {\mathrm{P}}_{\mathrm{t}}=26.25\hspace{0.33em}\mathrm{W} \end{gathered}$$

$$\begin{gathered} \mathrm{P}=\mathrm{\epsilon I} \\ \mathrm{P}=7.5\mathrm{V}\times 3.5\mathrm{A} \\ \mathrm{P}=26.25\mathrm{W} \end{gathered}$$

Also, we could confirm the energy conservation by getting the power output of the battery which in the case of internal resistance is zero it is given by,

Therefore the total power of the battery is 26.25 W