Circuit Power - AP Physics 2

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Question

Consider the circuit:

What is the rate of power consumption in the circuit if every resistor has a resistance of $3\Omega$ ?

Answer

To calculate the power consumption of the circuit, we need to first reduce it to an equivalent circuit with a single resistor. Since each resistor has the same resistance, this solution will keep resistance calculations as multiples of until the circuit is fully reduced.

Start with the two branches in parallel. We can condense R3 and R4, then solve for the total resistance of R2, R3, and R4.

$R_{34}=R+R=2R$

$\frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{R_2}+ \frac{1}{R_{34}}=\frac{1}{R}+\frac{1}{2R} = \frac{3}{2R}$

$R_{eq} = \frac{2}{3}R$

The equivalent circuit now has three resistors in series (R1, Req, and R5), so we can simply add them all up:

$R_{tot} = R+\frac{2}{3}R+R=\frac{8}{3}R$

Plug in the value for R:

$R_{tot}=\frac{8}{3}R=\frac{8}{3}(3\Omega)=8\Omega$

Now we can use the equation for power:

Substituting in Ohm's law for current, we get:
$P = \frac{V^2}{R} = \frac{(12V)^2}{8\Omega} = 18W$

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Question

Calculate the power consumed across resistor .

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

Answer

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

To calculate power, we need two of the following three quantities: voltage, current, and resistance.

In this case, since we are lacking the voltage, let's try to find the current.

We can use Kirchoff's junction law to calculate current .

The current coming into the junction = the current coming out of the junction.

Let's take a look at the central junction to the right of resistor .

$I_{1}=I_{2}+I_{3}$

$2=I_{3}+1$

$I_{3}=1$

Now that we know and , we can calculate power across the resistor.

$P={I^2}R=3W$

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Question

Elements A-D represent light bulbs.

Which of the following is true about these two circuits? Assume voltage sources have the same value and all the light bulbs are all identical.

Answer

Since bulbs A and B are in parallel, they will have the same voltage, and since the bulbs are identical in resistance, they will have the same current running through them and will be just as bright.

Let's say the voltage source as a value of and each bulb has a resistance of .

The current going through bulbs A and B is $\frac{V}{R}$ .

However, the current going through bulbs C and D is $\frac{V}{2R}$ .

The current going through bulbs C and D is half as much as the other two, so their brightness will be less.

So, bulbs A and B will be brigher than bulbs C and D.

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Question

You have 4 resistors, , , , and , set up like this:

Their resistance are as follows:

$\begin{align} A&=1\Omega \ B&=3\Omega \ C&=4\Omega \ D&= 2\Omega \end{align}$

If the battery has 8V, what is the total power dissipated through the resistors?

Answer

The equation for power is

In order to get the power, we need the current. To find the current, we need to get the total resistance, and use Ohm's Law ().

To find the total resistance, remember the equations for adding resistors is this:

$(Series)\ R_{tot}&=R_1+R_2+...+R_n$

$(Parallel)\ \frac{1}{R_{tot}}&=\frac{1}{R_1}+\frac{1}{R_2}=...=\frac{1}{R_n}$

Resistors and are in series, resistors and are in parallel, and resistors and are in series.

$R_{AB}&=1+3$

$R_{AB}= 4$

$R_{ABC}&=(\frac{1}{R_{AB}}+\frac{1}{R_C})^{-1}$

$R_{ABC}= (\frac{1}{4}+\frac{1}{4})^{-1}$

$R_{ABC}= 2$

$R_{total}&=R_{ABC}+R_{D}$

$R_{total}= 2+2$

$R_{total}= 4$

Now, we can find the current.

$I&=\frac{V}{R}$

$I=\frac{8}{4}$

Finally, we can find the power.

Therefore, the power is 16W (watts).

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Question

$R_A=2\Omega$

$R_B=4\Omega$

$R_C=6\Omega$

If the circuit above is connected to a battery, what is the total power dissipated by the circuit?

Answer

The equation for power dissipated in a circuit is

The three resistors are in parallel with each other, so the total resistance is

$(R_{tot})^{-1}=\left(\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}$

$(R_{tot})^{-1}=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)^{-1}$

$R_{tot}= \frac{8}{7}\Omega$

Use Ohm's law to find current.

$I&=\frac{V}{R}$

$I=\frac{4}{\frac{8}{7}}$

Finally, solve for power.

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Question

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the power being dissipated by .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}$

$\frac{3}{6}=\frac{1}{R_{1+2}}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

The current through and needs to add up to the total current, since they are in parallel.

$I_1 + I_2 =I_{total}$

$I_1 + I_2 =\frac{9}{8}A$

Also, the voltage drop across them need to be equal, since they are in parallel.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

Set up a system of equations.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

$I_1 + I_2 =\frac{9}{8}A$

Solve.

$1.5I_1=\frac{9}{8}A$

$I_1=\frac{3}{4}A$

The equation for power is as follows:

$P=\left(\frac{3}{4}A \right )^2\cdot3\Omega$

$P=\frac{27}{16}W$

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Question

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the power being dissipated by .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}$

$\frac{3}{6}=\frac{1}{R_{1+2}}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

The current through and needs to add up to the total current, since they are in parallel.

$I_1 + I_2 =I_{total}$

$I_1 + I_2 =\frac{9}{8}A$

Also, the voltage drop across them need to be equal, since they are in parallel.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

Set up a system of equations.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

$I_1 + I_2 =\frac{9}{8}A$

Solve.

$3I_2=\frac{9}{8}A$

$I_2=\frac{3}{8}A$

The equation for power is as follows:

$P=\left(\frac{3}{8}A \right )^2\cdot6\Omega$

$P=\frac{27}{32}W$

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Question

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the power being dissipaited by .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}$

$\frac{3}{6}=\frac{1}{R_{1+2}}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

Because it is not in parallel, the total current in the circuit is equal to the current in .

The equation for power is as follows:

$P=\left(\frac{9}{8}A \right )^2\cdot3\Omega$

$P=\frac{243}{64}W$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

Answer

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combining with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

$\frac{V}{R}=I$

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

The voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Using ohms law:

It is also true that:

$I_1+I_2=I_{Total}$

Using Subsitution

$I_1(1+\frac{R_1}{R_2})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_1}{R_2})}=I_1$

$\frac{3.43}{(1+\frac{1}{2})}=I_1$

Using the definition of electrical power, where is current and is the resistance of the component in question:

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Question

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

Answer

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

$\frac{V}{R}=I$

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

The voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Using ohms law:

It is also true that:

$I_1+I_2=I_{Total}$

Using Subsitution

$I_2(1+\frac{R_2}{R_1})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_2}{R_1})}=I_2$

Plugging in values

$\frac{1.14}{(1+\frac{2}{1})}=I_2$

Using the definition of electrical power, where is current and is the resistance of the component in question:

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Question

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

Answer

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

$\frac{V}{R}=I$

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Using ohms law:

It is also true that:

$I_3+I_4=I_{Total}$

Using Subsitution:

$I_3(1+\frac{R_3}{R_4})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_3}{R_4})}=I_3$

Plugging in values:

$\frac{3.43}{(1+\frac{3}{1})}=I_3$

Using the definition of electric power, where is current and is the resistance of the component in question.

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Question

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

Answer

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

$\frac{V}{R}=I$

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

It is true that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Using ohms law:

It is also true that:

$I_3+I_4=I_{Total}$

Using Subsitution:

$I_4(1+\frac{R_4}{R_3})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_4}{R_3})}=I_4$

Pluggin in values:

$\frac{3.43}{(1+\frac{1}{3})}=I_4$

Using the definition of electric power, where is current and is the resistance of the component in question.

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Question

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

Answer

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

$\frac{V}{R}=I$

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

It is true that the voltage drop across parallel resistors must be the same, so:

$V_{R5}=V_{R6}$

Using ohms law

It is also true that:

$I_5+I_6=I_{Total}$

Using Subsitution:

$I_5(1+\frac{R_5}{R_6})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_5}{R_6})}=I_5$

Plugging in values:

$\frac{3.43}{(1+\frac{2}{3})}=I_5$

Using the definition of electrical power, where is current and is the resistance of the component in question.

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Question

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

Answer

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

$\frac{V}{R}=I$

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

It is true that the voltage drop across parallel resistors must be the same, so:

$V_{R5}=V_{R6}$

Using ohms law

It is also true that:

$I_5+I_6=I_{Total}$

Using Subsitution:

$I_6(1+\frac{R_6}{R_5})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_6}{R_5})}=I_6$

Pluggin in values:

$\frac{3.43}{(1+\frac{3}{5})}=I_6$

Using the definition of electrical power, where is current and is the resistance of the component in question:

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5 \Omega$

$R_3=6 \Omega$

What is the power being dissapaited by ?

Answer

, , and are in parallel, so they are added by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

Plugging in given values:

$\frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{1}{R_{1,2,3}}}$

$R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So they are added conventionally:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

Plugging in values:

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, it is necessary to find the total current of the circuit. Using Ohm's law:

Solving for $\I$ :

$I=\frac{V}{R}$

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because , and are in parallel,

$I_1+I_2+I_3=I_{total}$

Also, the voltage drop must be the same across all three since they are in parallel.

Using Ohm's law again and substituting:

Using algebraic subsitution:

$I_1+I_1 \frac{R_1}{R_2}+I_1 \frac{R_1}{R_3}=I_{total}$

Solving for

$\frac{I_{Total}}{1+\frac{R_1}{R_2}+\frac{R_1}{R_3}}=I_1$

Plugging in values

$\frac{195/67}{1+\frac{2}{5}+\frac{2}{6}}=I_1$

Using the definition of electrical power, where is current and is the resistance of the component in question:

Plugging in values

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the power being dissapaited by ?

Answer

, , and are in parallel, so they are added by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

Plugging in given values:

$\frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{1}{R_{1,2,3}}}$

$R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So they are added conventionally:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

Plugging in values:

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, it is necessary to find the total current of the circuit. Using Ohm's law:

Solving for $\I$ :

$I=\frac{V}{R}$

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because , and are in parallel,

$I_1+I_2+I_3=I_{total}$

Also, the voltage drop must be the same across all three since they are in parallel.

Using Ohm's law again and substituting:

Using algebraic subsitution:

$I_2+I_2 \frac{R_2}{R_1}+I_2 \frac{R_2}{R_3}=I_{total}$

Solving for

$\frac{I_{total}}{1+\frac{R_2}{R_1}+\frac{R_2}{R_3}}=I_2$

Plugging in values:

$\frac{195/67}{1+\frac{5}{2}+\frac{5}{6}}=I_2$

Using the definition of electrical power, where is the current and is the resistance of the component in question:

$P=(.672amps)^2*5\Omega$

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the power being dissapaited by ?

Answer

, , and are in parallel, so they are added by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

Plugging in given values:

$\frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{1}{R_{1,2,3}}}$

$R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So they are added conventionally:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

Plugging in values:

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, it is necessary to find the total current of the circuit. Using Ohm's law:

Solving for $\I$ :

$I=\frac{V}{R}$

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because , and are in parallel,

$I_1+I_2+I_3=I_{total}$

Also, the voltage drop must be the same across all three since they are in parallel.

Using Ohm's law again and substituting:

Using algebraic subsitution:

$I_3+I_3 \frac{R_3}{R_1}+I_3 \frac{R_3}{R_2}=I_{total}$

Solving for :

$\frac{I_{Total}}{1+\frac{R_3}{R_1}+\frac{R_3}{R_2}}=I_3$

Plugging in values:

$\frac{195/67}{1+\frac{6}{2}+\frac{6}{5}}=I_3$

Using the definition of electric power, where is current and is reistance.

$P=(.560amps)^2*6\Omega$

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the power being dissapaited by ?

Answer

, , and are in parallel, so they are added by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

Plugging in given values:

$\frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{1}{R_{1,2,3}}}$

$R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So they are added conventionally:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

Plugging in values:

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, it is necessary to find the total current of the circuit. Using Ohm's law:

Solving for $\I$ :

$I=\frac{V}{R}$

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

The total current of the circuit is also the current through

Using the definition of electric power, where is current and is the resistance of the component in question:

$P=(\frac{195}{67})^2*2\Omega$

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What can be said about the power being dissipated by and ?

Answer

and are in series with each other, therefore, they will have the same current values. They also have the same resistance values. Thus, they will have the same power dissipated, as .

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Question

A single $9\textup{ V}$ battery is in series with several Ohmic resistors. How will the power output of the circuit change if a second $9\textup{ V}$ battery is added in series?

Answer

The power dissipated by the resistor will be

Using Ohm's Law:

And changing the equation to be exclusively in terms of voltage and resistance:

$P=\frac{V^2}{R}$

From this, it can be seen that doubling the voltage will quadruple the power.

$P=\frac{(2V)^2}{R}$

$P=\frac{4V^2}{R}$

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