All Solutions
Page 605: Section Review
We can start by writing down known equation for power:
$$P=IV$$
Using Ohm’s Law:
$$I=frac{V}{R}$$
and inserting it in the previous equation we get:
$$P=frac{V^2}{R}$$
The lower the resistance, the higher the power.
This finally means that:
$$boxed{R_{hot}<R_{warm}}$$
$$
P = IV
$$
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:
$$
boxed{ R_{hot} < R_{warm} }
$$
R_{hot} < R_{warm}
$$
$$
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}
$$
$V_f$ is the final voltage in the circuit and $P_f$ is the final power output in the circuit.
We can apply equation $(1)$ to the two cases above and we find that $P_i$ is equal to:
$$
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} }
$$
P_f = dfrac{P_i}{4}
$$
Power in the circuit decreased to a quarter of its initial value.
Power can be calculated as:
$$P=VI$$
Since compared appliances have equal power, we can see that higher voltage requires **lower currency**.
$$P_{losses}=RI^2$$
Which means that by lowering current, we are **significantly lowering** energy losses.
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.
textit{color{#c34632}$See$ $Explanation$}
$$
