Home Page Textbooks Physics: Principles and Problems Page 605: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

All Solutions

Page 605: Section Review

Exercise 32

Step 1

1 of 2

Car engine turns mechanical energy to electric energy which is then used by the generator. Generator uses this same energy and stores it to car’s battery as chemical energy. This chemical energy from the battery is then used in headlamps and part of the energy is used as energy of light in the headlamps, while the rest of the energy is turned to heat (thermal energy) in the headlamps and any other resistor.

Result

2 of 2

Hint: There is mechanical, electric, chemical, light and thermal energy.

Exercise 33

Solution 1

Solution 2

Step 1

1 of 5

We need to determine whether the resistance in the hair dryer will be smaller in hot or warm setting.

Step 2

2 of 5

In order to determine which setting has higher resitance, we can compare it through the only varible known to us, the power output; we know that the hot setting gives off higher power output than the warm setting.
We can start by writing down known equation for power:
$$P=IV$$

Step 3

3 of 5

Since current is unknown, resistance is yet to be determined, and voltage being constant, the easiest thing is to express the equation without current, we can do so by substituting current in the previous equation.
Using Ohm’s Law:
$$I=frac{V}{R}$$
and inserting it in the previous equation we get:
$$P=frac{V^2}{R}$$

Step 4

4 of 5

Since we already determined that the $P_{hot}>P_{warm}$, we can determine the resistance by looking at how resistance affects power.
The lower the resistance, the higher the power.
This finally means that:
$$boxed{R_{hot}<R_{warm}}$$

Result

5 of 5

$$R_{hot}<R_{warm}$$

Step 1

1 of 2

Notice that power output on the heater in the hair dryer is unknown in either of the cases, but we can find it from the definition of electric power

$$
P = IV
$$

where $I$ is the current flowing through the heater. Note that this current can be expressed from Ohm’s law, which states that the current $I$ flowing through a component with resistance $R$ and voltage $V$ across it is equal to:

$$
I = dfrac{V}{R }
$$

We’ll plug in current $I$ from the equation above into an equation for power and have:

$$
P = dfrac{V}{R} V = dfrac{V^2}{R}
$$

As we can see, power given off as heat in an arbitrary time interval is inversely proportional to the resistance $R$ used. Of course, hot setting on the hair dryer will give off more heat in the same time interval than the warm setting, which means that power $P_{hot}$ used in hot setting is greater than power $P_{warm}$ used in warm setting:

$$
P_{hot} > P_{warm}
$$

As said, power $P$ is inversely proportional to the resistance $R$, which means that the lower the resistance $R$, the greater the power output. From this we conclude that hot setting has lower resistance:

$$
boxed{ R_{hot} < R_{warm} }
$$

Result

2 of 2

$$
R_{hot} < R_{warm}
$$

Exercise 34

Step 1

1 of 3

Power output $P$ on an electric component is found as a product of current $I$ flowing through the component and voltage $V$ across it

$$
P = IV
$$

Note that this current can be expressed from Ohm’s law, which states that the current $I$ flowing through a component with resistance $R$ and voltage $V$ across it is equal to:

$$
I = dfrac{V}{R }
$$

We’ll plug in current $I$ from the equation above into an equation for power and have:

$$
begin{equation}
P = dfrac{V}{R} V = dfrac{V^2}{R}
end{equation}
$$

Step 2

2 of 3

Let $V_i$ be the initial voltage in the circuit and $P_i$ be the initial power output in the circuit, whereas
$V_f$ is the final voltage in the circuit and $P_f$ is the final power output in the circuit.
We can apply equation $(1)$ to the two cases above and we find that $P_i$ is equal to:

$$
begin{align*}
P_i = dfrac{V_i^2}{R} tag{2}
end{align*}
$$

whereas $P_f$ is equal to:

$$
begin{align*}
P_f = dfrac{V_f^2}{R} tag{3}
end{align*}
$$

We can use the fact that voltage in the circuit halved, which means that
$$
V_f = dfrac{V_i}{2}
$$
and divide equations $(3)$ and $(2)$:

$$
begin{align*}
dfrac{P_f}{P_i} &= dfrac{dfrac{V_f^2}{R} }{dfrac{V_i^2}{R}} \
tag{cancel out $R$} \
dfrac{P_f}{P_i} &= dfrac{V_f^2}{V_i^2} \
dfrac{P_f}{P_i} &= bigg( dfrac{V_f }{V_i}bigg)^2 \
tag{use the fact $V_f = dfrac{V_i}{2} $} \
dfrac{P_f}{P_i} &= left( dfrac{dfrac{V_i}{2} }{V_i}right)^2 \
tag{cancel out $V_i$} \
dfrac{P_f}{P_i} &= left( dfrac{1}{2} right)^2 \
dfrac{P_f}{P_i} &= dfrac{1}{4}
end{align*}
$$

$$
boxed{ P_f = dfrac{P_i}{4} }
$$

Result

3 of 3

$$
P_f = dfrac{P_i}{4}
$$

Power in the circuit decreased to a quarter of its initial value.

Exercise 36

Solution 1

Solution 2

Step 1

1 of 3

We need to determine the benefits of an electric water heater and electric range being connected to the $V_2=240,,rm{V}$ voltage source rather than $V_1=120,,rm{V}$

Step 2

2 of 3

In order to solve this problem, we need to take a look at the power and potential losses of energy.
Power can be calculated as:
$$P=VI$$
Since compared appliances have equal power, we can see that higher voltage requires **lower currency**.

Step 3

3 of 3

Lower currency is important because it is directly connected to the losses of electric energy:
$$P_{losses}=RI^2$$
Which means that by lowering current, we are **significantly lowering** energy losses.

Step 1

1 of 2

To get same power, at twice the voltage,
the current would be halved. The
I 2R loss in the circuit wiring would be
dramatically reduced since it is proportional
to the square of the current.

Result

2 of 2

$$
textit{color{#c34632}$See$ $Explanation$}
$$

Exercise 37

Step 1

1 of 2

We need to determine the positive effects of a brown-out.

Step 2

2 of 2

The power output of any resistive device, such as an electric space heater is equal to the power consumption, which is directly proportional to the voltage it is receiving. If the resistance stays constant, power consumption is proportional to the square of the applied voltage. Therefore, a significant reduction of **power output** will occur with a relatively small reduction in voltage.

unlock

All Solutions

Page 605: Section Review

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards