Chapter 18, Problem 6P

(a)

To determine

The resistance of a hair dryer, when it is plugged into a $\text{120}\text{V}$ line with $\text{7}\text{.5}\text{A}$ current.

Answer to Problem 6P

  $\text{16}\text{Ω}$

Explanation of Solution

Given:

Voltage, $\text{V=}120\text{V}$ current, $\text{I=}\text{7}\text{.5}\text{A}$

Formula used:

From the Ohm’s law

The relation between resistance, voltage and current

  $\text{V=IR}$

Where $V=$ voltage, $I=$ current, and $R=$ resistance

After rearranging the expression: $\text{R=}\frac{\text{V}}{\text{I}}$

Calculation:

Substituting the values in formula, $\text{R=}\frac{\text{V}}{\text{I}}$

  $\text{=}\frac{\text{120}\text{V}}{\text{7}\text{.5}\text{A}}$

  $\text{=16}\text{Ω}$

Conclusion:

The resistance of a hair dryer is $\text{16}\text{Ω}$ .

(b)

To determine

The charge passes through dryer in $\text{15}\text{min}$ .

Answer to Problem 6P

  $\text{6}{\text{.8×10}}^{\text{3}}\text{C}$

Explanation of Solution

Given:

Voltage, $\text{V=}120\text{V}$ , current, $\text{I=}\text{7}\text{.5}\text{A}$ and time, $\text{t = 15 min}$

Formula used:

The current is given by

  $\text{I}\text{=}\frac{\Delta \text{Q}}{\Delta \text{t}}$

For the calculation of charge $\Delta \text{Q}\text{=}\text{I}\Delta \text{t}$

Where, $I=$ current, $\Delta Q=$ amount of charge (change), and $\Delta t=$ change in time.

Calculation:

Substituting the values in formula $\Delta \text{Q}\text{=}\text{I}\Delta \text{t}$

  $\text{= 7}\text{.5}\text{A}\times 15\min \times 60\text{s/}\text{min}$

  $\text{ = 6}{\text{.8×10}}^{\text{3}}\text{C}$

Conclusion: The charge: $\text{6}{\text{.8×10}}^{\text{3}}\text{C}$