Chapter 22, Problem 97A

To determine

Range of the potentiometer.

Answer to Problem 97A

  $10\Omega \text{to }600\Omega$

Explanation of Solution

Given:

Voltage rating of motor, $V=12\text{V}$

Current in slowest setting, $I_1=0.02\text{A}$

Current in highest setting, $I_2=1.2\text{A}$

Formula Used:

Ohm’s law is given as,

  $V=IR$

Where, $V$ is voltage, $I$ is current and $R$ is resistance.

Calculation:

At slowest setting of motor,

  $\begin{array}{l}V=I_1{R}_1\\ 12=0.02\times R_1\\ \Rightarrow R_1=\frac{12}{0.02}\\ =600\Omega \end{array}$

At highest setting of motor,

  $\begin{array}{l}V=I_1{R}_1\\ 12=1.2\times R_2\\ \Rightarrow R_2=\frac{12}{1.2}\\ =10\Omega \end{array}$

So the range is $10\Omega \text{to }600\Omega$

Conclusion:

Therefore, the range of the potentiometer is $R_2\text{to }{R}_1\text{that is }10\Omega \text{to }600\Omega$ .