Chapter 18, Problem 10P

(a)

To determine

The current when voltage droops by $\text{15\%}$ .

Answer to Problem 10P

  $\text{5}\text{.5}\text{A}$

Explanation of Solution

Given:

Current, $\text{I=6}\text{.50A}$ voltage, $\text{V}\text{=240V}$ and decrease value of voltage is $\text{15\%}$ .

Formula used:

From the Ohm’s law,

The relation between resistance, voltage and current

  $\text{V=IR}$

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

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

Calculation:

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

Current $\text{I=6}\text{.50A}$ voltage $\text{V}\text{=240V}$ and decrease value of voltage is $\text{15\%}$

Relation between resistance, voltage and current -

  $\text{R=}\frac{\text{V}}{\text{I}}$

  $\text{=}\frac{\text{240}\text{V}}{\text{6}\text{.50}\text{A}}$

  $\text{=36}\text{.92}\text{Ω}$

After voltage $\text{15\%}$ decrease then

Voltage ${\text{V}}_{\text{new}}$

  $\text{=240-0}\text{.15×240}$

  $\text{=240-36}$

  $\text{=204}\text{volt}$

Calculation for the current, $\text{I=}\frac{{\text{V}}_{\text{new}}}{\text{R}}$

  $\text{=}\frac{\text{204}\text{V}}{\text{36}\text{.92}\Omega }\text{=5}\text{.5}\text{A}$

Conclusion:

The current after $\text{15\%}$ voltage decreases: $\text{5}\text{.5}\text{A}$

(b)

To determine

The current when resistance of device is reduced $\text{15\%}$ .

Answer to Problem 10P

  $\text{7}\text{.64}\text{A}$

Explanation of Solution

Given:

Current $\text{I=6}\text{.50A}$ voltage $\text{V}\text{=240V}$ and decrease value of resistance is $\text{15\%}$ .

Formula used:

From the Ohm’s law,

The relation between resistance, voltage and current

  $\text{V=IR}$

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

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

Calculation:

Substituting the values and we get

Current $\text{I=6}\text{.50A}$ voltage $\text{V}\text{=240V}$ and decrease value of resistance is $\text{15\%}$

Relation between resistance, voltage and current -

  $\text{R=}\frac{\text{V}}{\text{I}}$

  $\text{=}\frac{\text{240}\text{V}}{\text{6}\text{.50}\text{A}}$

  $\text{=36}\text{.92}\text{Ω}$

After the resistance is decrease $\text{15\%}$ then

  ${\text{R}}_{\text{new}}\text{=36}\text{.92-0}\text{.15}\times \text{36}\text{.92}$

  $\text{=}\text{36}\text{.92-5}\text{.538}$

  $\text{=31}\text{.382}\Omega$

Calculation for the current,

  $\text{I=}\frac{\text{V}}{{\text{ R}}_{\text{new}}}$

  $\text{=}\frac{\text{204}\text{V}}{\text{31}\text{.382}\Omega }\text{=7}\text{.64}\text{A}$

Conclusion:

The current after $\text{15\%}$ decrease in resistance is $\text{7}\text{.64}\text{A}$ .