In Fig. 27-55a, resistor 3 is a variable resistor and the ideal battery has emf.$\mathbf{}\mathit{\epsilon }\mathbf{=}\mathbf{}\mathbf{12}\mathbf{}\mathbf{\text{V}}$ Figure 27-55b gives the current I through the battery as a function of ${\mathit{R}}_{\mathbf{3}}$. The horizontal scale is set by.${\mathit{R}}_{\mathbf{3}\mathbf{s}}\mathbf{}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{\text{Ω}}$The curve has an asymptote of$\mathbf{2}\mathbf{.0}\mathbf{\text{\hspace{0.17em}mA}}$as${\mathit{R}}_{\mathbf{3}}\mathbf{}\mathbf{\to }\mathit{\infty }$. What are (a) resistance${\mathit{R}}_{\mathbf{1}}$and (b) resistance ${\mathit{R}}_{\mathbf{2}}$?

Emf of the ideal battery,$\epsilon =12\text{\hspace{0.17em}V}$

The horizontal scale is set by,${\text{R}}_{\text{3s}}=20\text{\hspace{0.17em}Ω}$

Current value at ,${\text{R}}_{\text{3}}\to \infty$,$\text{I}=2.0\text{\hspace{0.17em}mA}$

Using the current resistance relation based on Ohm's law, we can get that for a constant emf source or a constant voltage source, the current flowing through a variable resistor is inversely proportional to the values of the resistances of the resistor.

Formula:

The voltage equation using Ohm’s law, $\text{V}=\text{IR}$ (1)

When${\text{R}}_{\text{3}}=0$, all the current passes through${\text{R}}_{\text{1}}$and${\text{R}}_{\text{3}}$,avoids${\text{R}}_{\text{2}}$altogether.

Since that value of the current (through the battery) is$0.006\text{A}$(see Fig. 27-55(b)) for, ${\text{R}}_{\text{3}}=0$then using equation (1), the resistance can be given as follows:

$\begin{array}{c}{\text{R}}_{\text{1}}=\frac{\left(12\text{V}\right)}{\left(0.006\text{A}\right)}\\ =2.0\times {10}^3\text{\hspace{0.17em}Ω}\end{array}$

Hence, the value of the resistance is.$2.0\times {10}^3\text{\hspace{0.17em}Ω}$

When,${\text{R}}_{\text{3}}=\infty$all the current passes through${\text{R}}_{\text{1}}$and,${\text{R}}_{\text{3}}$avoids ${\text{R}}_{\text{3}}$altogether. Since that value of the current (through the battery) is 0.002 A (stated in the problem) for,${\text{R}}_{\text{3}}=\infty$then using equation (1), the resistance can be given as follows:

$\begin{array}{c}{\text{R}}_{\text{1}}=\frac{\left(12\text{V}\right)}{\left(0.002\text{A}\right)}\\ =6.0\times {10}^3\text{\hspace{0.17em}Ω}\end{array}$$\begin{array}{c}{\text{R}}_{\text{1}}=\frac{\left(12\text{V}\right)}{\left(0.002\text{A}\right)}\\ =6.0\times {10}^3\text{\hspace{0.17em}Ω}\end{array}$

Hence, the value of the resistance is.$6.0\times {10}^3\text{\hspace{0.17em}Ω}$