91Ó°ÊÓ

Emf of the ideal battery,$\mathrm{Î\mu }=12\text{ V}$

The horizontal scale is set by,${\text{R}}_{\text{3s}}=20\text{‰ө}$

Current value at ,${\text{R}}_{\text{3}}→\mathrm{∞}$,$\text{I}=2.0\text{″¾´¡}$

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:

$\begin{array}{c}{\text{R}}_{\text{1}}=\frac{\left(12\text{​â¶Ä‹V}\right)}{\left(0.006\text{A}\right)}\\ =2.0×{10}^3\text{‰ө}\end{array}$

Hence, the value of the resistance is.$2.0×{10}^3\text{‰ө}$

When,${\text{R}}_{\text{3}}=\mathrm{∞}$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,${\text{R}}_{\text{3}}=\mathrm{∞}$then using equation (1), the resistance can be given as follows:

$\begin{array}{c}{\text{R}}_{\text{1}}=\frac{\left(12\text{​â¶Ä‹V}\right)}{\left(0.002\text{A}\right)}\\ =6.0×{10}^3\text{‰ө}\end{array}$$\begin{array}{c}{\text{R}}_{\text{1}}=\frac{\left(12\text{​â¶Ä‹V}\right)}{\left(0.002\text{A}\right)}\\ =6.0×{10}^3\text{‰ө}\end{array}$

Hence, the value of the resistance is.$6.0×{10}^3\text{‰ө}$