Chapter 12.8, Problem 14P

a.

To determine

To calculate: What is the effect on the power. The power in an electric circuit varies directly as the square of the voltage and inversely as the resistance.

The resistance is constant and the voltage is halved.

Answer to Problem 14P

The statement is written as: “ $P=\frac{V^2}{4R}$ ”is $\frac{1}{4}$ times.

Explanation of Solution

Given information:

What is the effect on the power. The power in an electric circuit varies directly as the square of the voltage and inversely as the resistance.

The resistance is constant and the voltage is halved.

Formula used:

For the varies jointly and inversaly:

  $P=K\frac{V^2}{R}$

Calculation:

Consider the provided statement: What is the effect on the power. The power in an electric circuit varies directly as the square of the voltage and inversely as the resistance.

The resistance is constant and the voltage is halved.

Then rewrite the statement as:

Resistance $R$ is unchanged but $V$ is halved written as:

  $\begin{array}{l}P=\frac{{\left(\frac{V}{2}\right)}^2}{R}\\ P=\frac{V^2}{4R}\end{array}$ .

Here the power equals to the resistance is same but the voltage is halved.

Thus, the statement is written as: “ $P=\frac{V^2}{4R}$ ” $\frac{1}{4}$ times.

b.

To determine

To calculate: The voltage is constant but the resistance is halved.

Answer to Problem 14P

The expression “ $P=\frac{2V^2}{R}$ .” is $2$ times.

Explanation of Solution

Given information:

What is the effect on the power. The power in an electric circuit varies directly as the square of the voltage and inversely as the resistance.

The voltage is constant but the resistance is halved.

Formula used:

For the varies jointly and inversely:

  $P=K\frac{V^2}{R}$

Calculation:

Consider the provided statement “The voltage is constant but the resistance is halved.

Then rewrite the statement as:

Voltage $V$ is unchanged but resistance $I$ is halved written as:

  $\begin{array}{l}P=\frac{V^2}{\frac{R}{2}}\\ P=\frac{2V^2}{R}\end{array}$

Here the power is equals to the voltage is same but the resistance is halved.

Thus, the statement is written as: “ $P=\frac{2V^2}{R}$ ” $2$ times.

c.

To determine

To calculate: The voltage is tripled and the resistance is quadrupled.

Answer to Problem 14P

The statement is written as: “ $P=\frac{9}{4}\frac{V^2}{R}$ ” $\frac{9}{4}$ times.

Explanation of Solution

Given information:

The voltage is tripled and the resistance is quadrupled.

Formula used:

For the varies jointly and inversely:

  $P=K\frac{V^2}{R}$

Calculation:

Consider the provided statement The voltage is tripled and the resistance is quadrupled.

Then rewrite the statement as:

Resistance $R$ is quadrupled but voltage $V$ is tripled then the equation is written as:

  $\begin{array}{c}P=\frac{{\left(3V\right)}^2}{4R}\\ P=\frac{9V^2}{4R}\end{array}$ .

Here the power equals to the resistance is quadrupled but the voltage is tripled so it is $\frac{9}{4}$ times.

Thus, the statement is written as: “ $P=\frac{9}{4}\frac{V^2}{R}$ ” $\frac{9}{4}$ times.