Circuit Properties - AP Physics 2

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Question

Consider the three circuits shown. In each circuit, the voltage source is the same, and all resistors have the same value. If each resistor represents a light bulb, which of the three circuits will produce the brightest light?

Answer

To answer this question, we first need to determine what we're looking for that will allow any given light bulb to produce bright light. The answer is power. The more energy that is delivered to the light bulb within a given amount of time, the brighter the light will be. Let's go ahead and look at the equation for power in a circuit.

$Power= \frac{Energy}{Time}=i\Delta V$

And since we're told in the question stem that the voltage source is identical in each circuit, we're looking for the circuit that has the largest current.

In order to find which circuit has the largest current, we'll need to invoke Ohm's law.

$\Delta V=iR$

$i=\frac{\Delta V}{R}$

What this shows is that a higher current will occur in the circuit that has the lowest total resistance. Thus, we'll need to determine what the total resistance is in each of the three circuits.

In circuit A, we can see that there is only a single resistor. Thus, we can give this circuit's total resistance a value of .

In circuit B, we have two resistors that are connected in series. Remember that when resistors are connected in this way, the overall resistance of the circuit increases. We find the total resistance by summing all the resistors connected in series.

Therefore, we can give circuit B a total resistance of .

Now, let's look at circuit C. We can see that there are two resistors connected in parallel, and each of these are connected in series with a third resistor. To solve for the total resistance of this circuit, we first need to determine the equivalent resistance of the two resistors connected in parallel. Once we find that value, then we can take into account the third resistor connected in series.

To solve for the resistance of the two resistors connected in parallel, we have to remember that they add inversely.

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

$R_{Parallel}=\frac{R}{2}$

Now that we've found the equivalent resistance for the two resistors connected in parallel, we can consider the third resistor connected in series.

$R_{Total}=\frac{R}{2}+R=\frac{3R}{2}$

Thus, we can give circuit C a value of $\frac{3R}{2}$ .

Now that we've found the total resistance for each circuit, we can obtain our answer. Since circuit A has the lowest total resistance, it will also have the greatest current. Consequently, it will also have the greatest power delivered to its resistor (the light bulb), thus causing the light coming from that bulb to be the brightest.

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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

A single $250\ \textup{m}\Omega$ resistor is added in series to a circuit with a $9\textup{ V}$ battery. Determine the power dissipated by the resistor.

Answer

Use Ohm's law:

Converting $m\Omega$ to $\Omega$

Using definition of electric power:

Plugging in values:

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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:

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

Compare your answer with the correct one above

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