AP Physics 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?

Explanation

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).

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

None of these

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.

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:

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

None of these

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 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.

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

None of these

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 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.

A lone $3\Omega$ resistor is placed in series with a battery. How will adding a second $3\Omega$ resistor in series affect the power output of the circuit?

Cut it in half

Double it

Quadruple it

It will be unaffected

None of these

Explanation

Definition of electrical power:

Ohm's law:

Combining equations:

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

When adding resistors in series, they add directly

$R_1+R_2+.....=R_{total}$

Thus, in this case, resistance would increase, and if the voltage is kept constant, the power dissipated by the circuit would decrease. Doubling the resistance would cut the power output in half.

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

None of these

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 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.

Physics2set1q6

Calculate the power consumed across resistor .

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

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.

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

A single battery is in series with a $10\ \Omega$ resistor. How would the output of the circuit change if a second $10\ \Omega$ was added in series?

Halved

Quartered

Doubled

Tripled

Quadrupled

Explanation

The power dissipated by the resistor is

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 resistance will halve the power dissipated if the voltage is kept constant.

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Calculate the power being dissipated by

None of these

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 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:

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.

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

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.

Bulbs C and D have different brightnesses.

Bulbs A and B have different brightnesses.

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 .