All Solutions
Page 630: Practice Problems
$$P=P_1+P_2+P_3$$
$$P=2+3+1.5$$
$$P=6.5,,rm{W}$$
$$P=IV$$
where $V=epsilon$ and $I$ is a total current of the circuit, which is also the current supplied by the battery.
$$I=frac{P}{V}$$
$$I=frac{6.5}{12}$$
Finally:
$$boxed{I=0.54,,rm{A}}$$
$P_1 = 2 ~mathrm{W }$, second dissipates $P_2 = 3 ~mathrm{W }$ and the third resistor dissipates $P_3 = 1.5 ~mathrm{W }$. We know that total power $P$ dissipated on the resistors in a circuit with any type of connections between the resistors is equal to a sum of power dissipated on individual resistors, stated as:
$$
begin{align*}
P &= P_1 + P_2 + P_3 \
tag{plug in the values} \
P &= 2 ~mathrm{W } + 3 ~mathrm{W } + 1.5 ~mathrm{W } \
P &= 6.5 ~mathrm{W }
end{align*}
$$
Total power dissipation $P$ can also be calculated as a product of voltage $V$ across the battery and current $I$ supplied by the battery, stated as:
$$
begin{align*}
P &= I V \
tag{express $I$ from the equation above } \
I &= dfrac{P}{V} \
I &= dfrac{ 6.5 ~mathrm{W } }{12 ~mathrm{V } }
end{align*}
$$
$$
boxed{ I = 0.5416 ~mathrm{A}}
$$
I = 0.5416 ~mathrm{A}
$$
$$
E = P t
$$
As we can see, energy output to light is proportional to the power rating of the device. This means that we must decide which of the lights will have the higher power output if lights, connected in this way, are connected across an arbitrary source of voltage $V$.
$$
begin{align*}
I = dfrac{V} {R_e} tag{1}
end{align*}
$$
where $R_e$ is equivalent resistance of all the resistors in the circuit. We can find the equivalent resistance of resistors in this circuit by first finding the equivalent resistance $R_{parallel}$ of the two lights connected in parallel from:
$$
begin{align*}
dfrac{1}{R_{parallel}} &= dfrac{1}{R}+ dfrac{1}{R} \
dfrac{1}{R_{parallel}} &= dfrac{2}{R} \
R_{parallel} &= dfrac{R}{2}
end{align*}
$$
$$
begin{align*}
R_e &= 11 R + R_{parallel} \
R_e &= 11R + dfrac{R}{2} \
R_e &= 11.5 R
end{align*}
$$
We can now calculate current $I$ from equation $(1)$ as:
$$
begin{align*}
I &= dfrac{V} {R_e} \
I &= dfrac{V }{ 11.5 R }
end{align*}
$$
This same current $I$ will flow through the 11 lights connected in series. Power rating $P_{series}$ on these lights can be expressed as:
$$
begin{align*}
P_{series} &= I^2 R \
tag{plug in the calculated current $I$} \
P_{series} &= bigg( dfrac{V }{ 11.5 R } bigg)^2 \
P_{series} &= dfrac{V^2}{(11.5)^2 R^2 } R \
P_{series} &= dfrac{V^2}{132.25 R }
end{align*}
$$
$$
V = 11 V_{series} + V_{parallel}
$$
Voltage $V_{series}$ across each light connected in series can be obtained from Ohm’s law, which states that this voltage is equal to a product of current $I$ flowing through these lights and resistance $R$ across them, stated as:
$$
begin{align*}
V_{series} &= I R \
tag{plug in the calculated values} \
V_{series} &= dfrac{V }{ 11.5 R } R \
V_{series} &= dfrac{V}{11.5}
end{align*}
$$
We can plug this expression for $V_{series}$ into equation above to calculate $V_{parallel}$ as:
$$
begin{align*}
V &= 11 V_{series} + V_{parallel} \
V &= dfrac{11 V}{11.5} + V_{parallel} \
tag{express $V_{parallel}$ from the equation above} \
V_{parallel} &= V – dfrac{11V }{11.5} \
tag{group the terms in the brackets} \
V_{parallel} &= V bigg( 1- dfrac{11}{11.5} bigg) \
V_{parallel} &= V bigg( dfrac{11.5 – 11}{11.5} bigg) \
V_{parallel} &= V dfrac{0.5}{11.5} \
tag{replace $0.5 = dfrac{1}{2} $} \
V_{parallel} &= V dfrac{1}{2 cdot 11.5} \
V_{parallel} &= dfrac{V}{23}
end{align*}
$$
$$
begin{align*}
P_{parallel} &= dfrac{V_{parallel}^2}{R} \
tag{plug in $ V_{parallel} = dfrac{V}{23}$ } \
P_{parallel} &= bigg( dfrac{V}{23} bigg)^2 dfrac{1}{R} \
P_{parallel} &= dfrac{V^2}{23^2 R } \
P_{parallel} &= dfrac{V^2}{529 R}
end{align*}
$$
$$
begin{align*}
dfrac{P_{parallel}}{P_{series}} &= dfrac{dfrac{V^2}{529 R}}{dfrac{V^2}{132.25 R}} \
tag{cancel out $V^2$ and $R$} \
dfrac{P_{parallel}}{P_{series}} &= dfrac{132.25}{529} \
dfrac{P_{parallel}}{P_{series}} &= 0.25 \
tag{find reciprocal value of the equation above} \
dfrac{P_{series}}{P_{parallel}} &= dfrac{1}{0.25} \
tag{replace $0.25 = dfrac{1}{4}$} \
dfrac{P_{series}}{P_{parallel}} &= 4
end{align*}
$$
$$
boxed{ P_{series} = 4 P_{parallel} }
$$
$$
E = P t
$$
we conclude that the equation
$$
P_{series} = 4 P_{parallel}
$$
means that the lights connected in series will shine 4 times more brightly than the two lights connected in parallel.
P_{series} = 4 P_{parallel}
$$
Lights connected in series will shine 4 times more brightly than the two lights connected in parallel.
$$R_e=12cdot R$$
$$R_{e0}=11cdot R+frac{Rcdot R}{R+R}$$
Which is:
$$R_{e0}=11.5cdot R$$
Comparing the current before and after we get:
$$frac{I}{I_o}=frac{R_{e0}}{E_{e}}$$
Inserting values we get:
$$frac{I}{I_0}=frac{11.5}{12}$$
$$boxed{I=0.958cdot I_0}$$
Let’s say this circuit is connected to an arbitrary voltage source with voltage $V$.
If one of the two lights connected in parallel burns out, current won’t flow through it but it will keep flowing through the other light connected to it in parallel. This way, we essentially have 12 identical lights connected in series. This way, equivalent resistance $R_e$ of the 12 identical lights with resistance $R$ is calculated as:
$$
R_e = 12 R
$$
Current $I$ supplied by the battery can be calculated by dividing the voltage $V$ supplied by the battery and equivalent resistance of all the resistors in the circuit, stated as:
$$
begin{align*}
I &= dfrac{V}{R_e} \
I &= dfrac{V}{12 R}
end{align*}
$$
As we can see, back when one of the lights connected in parallel didn’t burn out, current was $I_0 = dfrac{V}{11.5 R}$. Clearly, if one of the lights connected in parallel burns out, current will keep flowing because it can continue flowing through the other branch containing the other light that didn’t burn out.
Let’s compare the two currents:
$$
begin{align*}
dfrac{I}{I_0} &= dfrac{dfrac{V}{12 R}}{dfrac{ V }{11.5 R}} \
dfrac{I}{I_0} &= dfrac{11.5}{12} \
dfrac{I}{I_0} &= 0.95833 \
I &= 0.95833 I_0
end{align*}
$$
As we can see, current decreased.
That means that the equivalent resistance can be calculated only as 11 bulbs in series:
$$R_e=11cdot R$$
$$R_{e0}=11cdot R + frac{Rcdot R}{R+R}$$
Which is:
$$R_{e0}=11.5cdot R$$
$$frac{I}{I_0}=frac{R_{e0}}{R_r}$$
$$frac{I}{I_0}=frac{11.5}{11}$$
Finally:
$$boxed{I=1.045cdot I_0}$$
Let’s say this circuit is connected to an arbitrary voltage source with voltage $V$.
If one of the two lights connected in parallel is shorted out, current won’t flow through either of them, but it will keep flowing through the wire with which we shorted them. The reason for this is due to a fact that current will always flow through a wire with less resistance, and resistance is zero through a wire with which we short any device. This way, we’ll only have the 11 identical lights connected in series, while the two connected in series wont’ light at all. This way, equivalent resistance $R_e$ of the 11 identical lights with resistance $R$ is calculated as:
$$
R_e = 11 R
$$
Current $I$ supplied by the battery can be calculated by dividing the voltage $V$ supplied by the battery and equivalent resistance of all the resistors in the circuit, stated as:
$$
begin{align*}
I &= dfrac{V}{R_e} \
I &= dfrac{V}{11 R}
end{align*}
$$
As we can see, back when one of the lights connected in parallel wasn’t shorted, current was $I_0 = dfrac{V}{11.5 R}$. Clearly, if one of the lights connected in parallel is shorted, current will keep flowing because it can continue flowing through the wire with which we made the short-circuit.
Let’s compare the two currents:
$$
begin{align*}
dfrac{I}{I_0} &= dfrac{dfrac{V}{11 R}}{dfrac{ V }{11.5 R}} \
dfrac{I}{I_0} &= dfrac{11.5}{11} \
dfrac{I}{I_0} &= 1.04545 \
I &= 1.04545 I_0
end{align*}
$$
As we can see, current increased.
