In Fig. E26.11, the battery has emf 35.0 V and negligible internal resistance. ${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathit{\Omega }$. The current through is ${\mathit{R}}_{\mathbf{1}}$, 1.5 A and the current through ${\mathit{R}}_{\mathbf{3}}$is 4.50 A. What are the resistances ${\mathit{R}}_{\mathbf{1}}$and ${\mathit{R}}_{\mathbf{3}}$?

Internal resistance refers to the opposition to the flow of charges offered by the cells and batteries themselves resulting in the generation of heat.

Given data:

• $\epsilon =35.0\hspace{0.33em}V$
• $I_3=4.5\hspace{0.33em}A$
  
• $I_1=1.5\hspace{0.33em}A$
  
• r = 0

As$R_1$ and$R_2$ are in parallel, this means they have the same voltage that equals to the voltage across their combination $R_{12}$. So, the voltage across$R_2$ is given using Ohm’s law as:

$V_2=V_1=I_1{R}_1$

Plug the values,

$$\begin{gathered} V_2=\left(1.5A\right)\left(5.0\Omega \right) \\ =7.5V \end{gathered}$$

Now, the combination is in series with , which means the current is the same for both as:

$I_3=I_{1}2=I_1+I_2$

Plug the values,

$$\begin{gathered} l_2=l_3-l_1 \\ =4.5-1.5 \\ =3.0A \end{gathered}$$

Now, use Ohm’s law to calculate resistance as:

$$\begin{gathered} R_2=\frac{V_2}{l_2} \\ =\frac{7.5}{3.0} \\ =2.5\Omega \end{gathered}$$

The voltage drop of the battery is due to $R_3$and the combination $R_{12}$. And as the internal resistance is zero, the Ohm's law can be used as:

$$\begin{gathered} \epsilon =I(R_{12}+R_3) \\ {R}_3=\frac{\epsilon }{l}=R_{1}2 \end{gathered}$$ (1)

The combination can be calculated as:

$$\begin{gathered} R_{12}=\frac{R_1{R}_2}{R_1+R_2} \\ =\frac{\left(5.0\right)\left(2.5\right)}{5.0+2.5} \\ =1.67\Omega \end{gathered}$$

Plug the values in equation (1),

$$\begin{gathered} R_3=\frac{35.0V}{4.5A}-1.67\Omega \\ =6.11\Omega \end{gathered}$$

Thus, the value of resistances is $R_2=2.5\Omega$and $R_3=6.11\Omega$.