All Solutions
Page 615: Standardized Test Practice
$$P=VI$$
$$I=frac{P}{V}$$
$$I=frac{100}{120}$$
Finally, the current is:
$$boxed{I=0.8,,rm{A}}$$
$$
begin{align*}
P &= VI \
tag{express $I$ from the equation above} \
I &= dfrac{P}{V} \
tag{plug in the given values} \
I &= dfrac{100 ~mathrm{W}}{120 ~mathrm{V}} \
I &= 0.8333 ~mathrm{A}
end{align*}
$$
$$
boxed{ A)~~ I = 0.8 ~mathrm{A} }
$$
A)~~ I = 0.8 ~mathrm{A}
$$
$$
E = Q = P t
$$
where $P$ is power output of the resistor and $t$ is the time for which current was flowing through the resistor. We see that in order to find the thermal energy released from the resistor over a time period of $t = 7.5 ~mathrm{min}$, we first need to find the power output $P$ of the resistor.
$$
E = P t
$$
where $P$ is power output of the resistor, which can be calculated as:
$$
P = dfrac{V^2}{R}
$$
We can plug in this expression for power output $P$ into the equation above and have:
$$
begin{align*}
E &= dfrac{V^2}{R} t \
tag{plug in the values} \
E &= dfrac{ ( 9 ~mathrm{V})^2 }{ 5 ~mathrm{Omega } } cdot 7.5 ~mathrm{min} \
tag{$ 1 ~mathrm{min} = 60 ~mathrm{s}$} \
E &= dfrac{ 81~mathrm{V}^2 }{ 5 ~mathrm{Omega } } cdot 7.5 cdot 60 ~mathrm{s} \
E &= 7290 ~mathrm{J} \
E &approx 7.3 cdot 10^3 ~mathrm{J}
end{align*}
$$
As said, all of the supplied electric energy $E$ is released as thermal energy $Q$, which means:
$$
boxed{ D)~~ Q approx 7.3 cdot 10^3 ~mathrm{J} }
$$
D)~~ Q approx 7.3 cdot 10^3 ~mathrm{J}
$$
$$
P = VI
$$
$$
begin{align*}
V_{total} &= V_{battery} + V_{battery} + V_{battery} \
V_{total} &= 3 V_{battery} \
tag{plug in the values} \
V_{total} &= 3 cdot 1.5 ~mathrm{V} \
V_{total} &= 4.5 ~mathrm{V}
end{align*}
$$
$$
V_{total} = V
$$
This means that the power $P$ delivered to the bulb can be calculated as:
$$
begin{align*}
P &= VI \
tag{plug in $V = V_{total}$} \
P &= V_{total} I \
tag{plug in the values} \
P &= 4.5 ~mathrm{V} cdot 0.5 ~mathrm{A} \
P &= 2.25 ~mathrm{W}
end{align*}
$$
In other words, power $P$ delivered to the bulb of the flashlight is approximately equal to:
$$
boxed{ C)~~ P approx 2.3 ~mathrm{W} }
$$
C)~~ P approx 2.3 ~mathrm{W}
$$
$$
begin{align*}
E &= P t tag{1}
end{align*}
$$
where $P$ is power rating of the device. We see that in order to find the electric energy supplied to the flashlight, we first need to find its power rating $P$, which can be calculated as:
$$
P = VI
$$
$$
begin{align*}
V_{total} &= V_{battery} + V_{battery} + V_{battery} \
V_{total} &= 3 V_{battery} \
tag{plug in the values} \
V_{total} &= 3 cdot 1.5 ~mathrm{V} \
V_{total} &= 4.5 ~mathrm{V}
end{align*}
$$
$$
V_{total} = V
$$
This means that the power $P$ delivered to the bulb can be calculated as:
$$
begin{align*}
P &= VI \
tag{plug in $V = V_{total}$} \
P &= V_{total} I \
tag{plug in the values} \
P &= 4.5 ~mathrm{V} cdot 0.5 ~mathrm{A} \
P &= 2.25 ~mathrm{W}
end{align*}
$$
$t = 3 ~mathrm{min}$ by using equation $(1)$:
$$
begin{align*}
E &= P t \
tag{plug in the values} \
E &= 2.25 ~mathrm{W} cdot 3 ~mathrm{min} \
tag{$ 1 ~mathrm{min } = 60 ~mathrm{s} $} \
E &= 2.25 ~mathrm{W} cdot 3 cdot 60 ~mathrm{s} \
E &= 405 ~mathrm{J} \
E &= 4.05 cdot 10^2 ~mathrm{J}
end{align*}
$$
We’ve calculated that the electric energy supplied to the flashlight is approximately equal to:
$$
boxed{ D)~~ E approx 4.1 cdot 10^2 ~mathrm{J} }
$$
D)~~ E approx 4.1 cdot 10^2 ~mathrm{J}
$$
We must find how much of the electric energy $E_{electric}$ supplied to the motor is converted to mechanical energy $E_{mechanical}$ over a time period of $t = 1 ~mathrm{min}$.
Since we have a current flowing through a motor, this means that the electric energy $E_{electric}$ supplied to the motor will be transformed to the mechanical energy $E_{mechanical}$ that the motor uses to operate. In other words, these two energies will be equal:
$$
E_{mechanical} = E_{electric}
$$
We can calculate the electric energy $E_{electric}$ supplied to the motor over time $t$ as:
$$
E_{electric} = P t
$$
where $P$ is electric power delivered to the motor. As said, this electric energy is equal to mechanical energy used by the motor:
$$
begin{align*}
E_{mechanical} &= P t tag{1}
end{align*}
$$
$$
P = I^2 R
$$
We can plug in this expression for electric power $P$ into the equation $(1)$ and calculate the mechanical energy used by the motor:
$$
begin{align*}
E_{mechanical} &= P t \
tag{plug in $P = I^2 R$} \
E_{mechanical} &= I^2 R t\
tag{plug in the values } \
E_{mechanical} &= (2 ~mathrm{A})^2 cdot 12 ~mathrm{Omega} cdot 1 ~mathrm{min} \
tag{$ 1 ~mathrm{min } = 60 ~mathrm{s} $} \
E_{mechanical} &= 4 ~mathrm{A} ^2 cdot 12 ~mathrm{Omega} cdot 60 ~mathrm{s}\
E_{mechanical} &= 2880 ~mathrm{J} \
E_{mechanical} &= 2.88 cdot 10^3 ~mathrm{J}
end{align*}
$$
We’ve found that the electric energy that gets converted to the mechanical energy is approximately equal to:
$$
boxed{ C)~~ E_{mechanical} approx 2.9 cdot 10^3 ~mathrm{J} }
$$
C)~~ E_{mechanical} approx 2.9 cdot 10^3 ~mathrm{J}
$$
$$
I = dfrac{V}{R}
$$
$$
R_f = dfrac{1}{2} R
$$
$$
text{and}
$$
$$
V_f = dfrac{1}{2} V
$$
where $R_f$ and $V_f$ are resistance and voltage in the final state of the circuit, current $I_f$ flowing through the circuit when we reduce both voltage and resistance by half will be equal to:
$$
I_f = dfrac{V_f}{R_f}
$$
To find if and by how much the current changed when both resistance and voltage are halved, we need to calculate the current $I_f$ in the final state.
$$
begin{align*}
I_f &= dfrac{V_f}{R_f} \
tag{plug in $V_f = dfrac{1}{2} V $ } \
tag{plug in $R_f = dfrac{1}{2} R $ } \
I_f &= dfrac{dfrac{1}{2} V }{dfrac{1}{2} R} \
tag{cancel out $dfrac{1}{2}$} \
I_f &= dfrac{V}{R} \
tag{notice $dfrac{V}{R} = I$ } \
I_f &= I
end{align*}
$$
As we can see, current flowing through the circuit didn’t change. This was expected since current $I$ is proportional to the voltage $V$ across the voltage source and inversely proportional to the resistance $R$ of the resistor. If both of the parameters are changed by the same amount, current $I$ won’t change. We thus conclude that the final answer is:
$$
boxed{ B)~~ text { No change } }
$$
B)~~ text { No change }
$$
$$P=VI$$
and Ohm’s Law:
$$V=RI$$
$$P=I^2R$$
$$P=(5cdot 10^{-3})^2cdot 50$$
Finally we get:
$$boxed{P=1.25cdot 10^{-3},,rm{W}}$$
$$
begin{align*}
P &= I^2 R \
tag{plug in the values} \
P &= (5 ~mathrm{mA})^2 cdot 50 ~mathrm{Omega} \
tag{$ 1 ~mathrm{mA} = 10^{-3} ~mathrm{A} $} \
P &= (5 cdot 10^{-3} ~mathrm{A} )^2 cdot 50 ~mathrm{Omega} \
P &= 25 cdot 10^{-6} ~mathrm{A} ^2 cdot 50 ~mathrm{Omega} \
P &= 1250 cdot 10^{-6} ~mathrm{W} \
P &= 1.25 cdot 10^{-3} ~mathrm{W}
end{align*}
$$
We found that the final answer is:
$$
boxed{ C)~~ P = 1.25 cdot 10^{-3} ~mathrm{W} }
$$
C)~~ P = 1.25 cdot 10^{-3} ~mathrm{W}
$$
$$
E = P t
$$
From the equation above we see that the greater the power rating $P$ of the lightbulb, the more electric energy it used. Also, the more time $t$ the lightbulb was on, the more electric energy it used.
$P = 60 ~mathrm{W}$ over a given time period of
$t = 2.5 ~mathrm{h}$. Let’s apply the equation above to this lightbulb:
$$
begin{align*}
E &= P t \
tag{plug in the values} \
E &= 60 ~mathrm{W} cdot 2.5 ~mathrm{h} \
tag{$ 1 ~mathrm{h} = 60 ~mathrm{min} = 60 cdot 60 ~mathrm{s} = 3600 ~mathrm{s} $} \
E &= 60 ~mathrm{W} cdot 2.5 cdot 3600 ~mathrm{s} \
E &= 540 000 ~mathrm{J} \
E &= 5.4 cdot 10^5 ~mathrm{J}
end{align*}
$$
We found that the final answer is:
$$
boxed{ D)~~ E = 5.4 cdot 10^5 ~mathrm{J} }
$$
D)~~ E = 5.4 cdot 10^5 ~mathrm{J}
$$
$$
begin{align*}
text{Resistance of the hair dryer}~ R_{hair~dryer} &= 8.5 ~mathrm{Omega } \
text{Resistance of the heater }~ R_h &= 10 ~mathrm{Omega } \
text{Resistance of the small motor}~ R_sm &= 12 ~mathrm{Omega } \
end{align*}
$$
As said, one of the three resistors given above will be used in the circuit. In our case, we’ll use resistance $R_{hair~dryer}$ of the hair dryer and we are asked to calculate the current $I$ flowing in the circuit and how much thermal energy is produced in a time period of $t = 2.5 ~mathrm{min}$.
Since this is a simple circuit, same current fill flow through all of the circuit and it can be calculated from the Ohm’s law as:
$$
I = dfrac{V}{R }
$$
where $R$ is resistance of the resistor through which the current $I$ flows.
Since we have a resistor connected to a battery, current $I$ will flow through the circuit, resistor will heat up and release thermal energy $Q$. Of course, the more the time $t$ the resistor heats up, the greater the release of thermal energy. We’ll assume that all of the electric energy supplied to the resistor is used as thermal energy, which can be calculated as:
$$
E = Q = P t
$$
where $P$ is power output of the resistor and $t$ is the time for which current was flowing through the resistor. We see that in order to find the thermal energy released from the resistor over a time period of $t = 2.5 ~mathrm{min}$, we first need to find the power output $P$ of the resistor.
$$
I = dfrac{V}{R }
$$
When we apply the Ohm’s law to our circuit, with unknown current $I$ flowing through resistor $R_{hair~dryer}$ with voltage $V$ across it, we have:
$$
begin{align*}
I &= dfrac{V}{R_{hair~dryer}} \
tag{plug in the values} \
I &= dfrac{120 ~mathrm{V}}{8.5 ~mathrm{Omega}}
end{align*}
$$
$$
boxed{ I = 14.1176 ~mathrm{A} }
$$
$$
E = P t
$$
where $t$ is time for which current was flowing through the resistor and
$P$ is power output of the resistor, which can be calculated as:
$$
P = dfrac{V^2}{R_{hair~dryer}}
$$
We can plug in this expression for power output $P$ into the equation above and have:
$$
begin{align*}
E &= dfrac{V^2}{R_{hair~dryer}} t \
tag{plug in the values} \
E &= dfrac{ ( 120 ~mathrm{V})^2 }{8.5 ~mathrm{Omega } } cdot 2.5 ~mathrm{min} \
tag{$ 1 ~mathrm{min} = 60 ~mathrm{s}$} \
E &= dfrac{ 14400~mathrm{V}^2 }{8.5 ~mathrm{Omega } } cdot 2.5 cdot 60 ~mathrm{s} \
E &= 254 117.6471 ~mathrm{J}
end{align*}
$$
As said, all of the supplied electric energy $E$ is released as thermal energy $Q$, which means:
$$
boxed{ Q = 254 117.6471 ~mathrm{J} }
$$
I = 14.1176 ~mathrm{A}
$$
$$
Q = 254 117.6471 ~mathrm{J}
$$
