Question

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 ${\mathit{R}}_{\mathbf{3}}$?

Short answer

The value of resistances is$R_2=2.5\Omega$ and $R_3=6.11\Omega$.

Step-by-step solution

Definition of Internal Resistance

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

Determine the resistance R2

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:

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

Determine the resistance R3

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 $R_3=6.11\Omega$.