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AP Physics 2 : Circuit Power

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Example Question #1 : Circuit Power

Consider the circuit:

Circuit_1

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

Possible Answers:

21 W

12 W

24 W

18W

15 W

Correct answer:

18W

Explanation:

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:

P = IV

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

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Example Question #2 : Circuit Power

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

4resistorcircuit

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?

Possible Answers:

16W

20W

12W

Correct answer:

16W

Explanation:

The equation for power is

P=VI

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 ( V = IR ).

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 ABC 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.

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

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

I= 2

Finally, we can find the power.

P&=VI

P=(8)(2)

P= 16

Therefore, the power is 16W (watts).

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Example Question #3 : Circuit Power

Physics2set1q6

Calculate the power consumed across resistor .

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

Possible Answers:

10W

3 W

Correct answer:

3 W

Explanation:

Physics2set1q6

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

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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Example Question #4 : Circuit Power

Physics2set1q4a Physics2set1q4b

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.

Possible Answers:

Bulbs A and B have different brightnesses.

Bulbs A and B will be brighter than bulbs C and D.

Bulbs C and D have different brightnesses.

All bulbs will have the same brightness.

Bulb A will be as bright as bulb C, but bulbs B and D will have a different brightness.

Correct answer:

Bulbs A and B will be brighter than bulbs C and D.

Explanation:

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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Example Question #1 : Circuit Power

Photo 4 1

$R_A=2\Omega$

$R_B=4\Omega$

$R_C=6\Omega$

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

Possible Answers:

$\frac{8}{7}W$

18W

$\frac{28}{11}W$

10W

14W

Correct answer:

14W

Explanation:

The equation for power dissipated in a circuit is

P=IV

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.

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

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

I= 3.5A

Finally, solve for power.

P=IV

P=(4)(3.5)

P= 14W

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Example Question #1 : Circuit Power

Combined circuit

V_1 = V_2 = 4.5V

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the power being dissipaited by R_3 .

Possible Answers:

$\frac{243}{64}W$

$\frac{9}{8}W$

$\frac{3}{4}W$

None of these

Correct answer:

$\frac{243}{64}W$

Explanation:

First, find the total resistance of the circuit.

R_1 and R_2 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.

V=IR

(V_1+V_2)=IR

$9V=I\cdot 8\Omega$

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

Because it is not in parallel, the total current in the circuit is equal to the current in R_3 .

The equation for power is as follows:

P=IV=I^2R

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

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

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Example Question #1 : Circuit Power

Combined circuit

V_1 = V_2 = 4.5V

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the power being dissipated by R_2 .

Possible Answers:

$\frac{27}{32}W$

$\frac{9}{8}W$

$\frac{27}{64}W$

None of these

Correct answer:

$\frac{27}{32}W$

Explanation:

First, find the total resistance of the circuit.

R_1 and R_2 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.

V=IR

(V_1+V_2)=IR

$9V=I\cdot 8\Omega$

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

The current through R_1 and R_2 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.

V_1=V_2

I_1R_1=I_2R_2

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

Set up a system of equations.

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

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

Solve.

I_2=.5I_1

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

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

The equation for power is as follows:

P=IV=I^2R

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

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

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Example Question #1 : Circuit Power

Combined circuit

V_1 = V_2 = 4.5V

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the power being dissipated by R_1 .

Possible Answers:

$\frac{9}{8}W$

$\frac{3}{4}W$

None of these

$\frac{27}{16}W$

Correct answer:

$\frac{27}{16}W$

Explanation:

First, find the total resistance of the circuit.

R_1 and R_2 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.

V=IR

(V_1+V_2)=IR

$9V=I\cdot 8\Omega$

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

The current through R_1 and R_2 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.

V_1=V_2

I_1R_1=I_2R_2

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

Set up a system of equations.

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

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

Solve.

I_2=.5I_1

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

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

The equation for power is as follows:

P=IV=I^2R

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

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

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Example Question #1 : Circuit Power

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What can be said about the power being dissipated by R_4 and R_5 ?

Possible Answers:

None of these

R_4 will dissipate more power than R_5

R_5 will dissipate more power than R_4

It is impossible to tell

Their values will be equal

Correct answer:

Their values will be equal

Explanation:

R_4 and R_5 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 P=I^2R .

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Example Question #10 : Circuit Power

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

V_1=V_2=V_3=3 V

Calculate the power being dissipated by R_3

Possible Answers:

2.21 watts

37.5 watts

1.55 watts

None of these

9.15 watts

Correct answer:

2.21 watts

Explanation:

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 R_1 with R_2 , R_3 with R_4 , R_5 with R_6 .

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

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

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

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

V=IR

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

Combing all voltage sources for the total voltage.

V_1+V_2+V_3=9 Volts

Plugging in given values,

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

I=3.43 amps

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

$V_{R3}=V_{R4}$

Using ohms law:

I_3R_3=I_4R_4

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 I_3 :

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

Plugging in values:

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

.858 amps=I_3

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

P=I^2R

P=.858^2*3

P= 2.21 watts

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AP Physics 2 : Circuit Power

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