Circuits - AP Physics 1

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Question

Consider the following circuit:

How much power is lost through R1?

$R1 = 5\Omega,\ R2 = 3\Omega,\ R3 = 4\Omega,\ R4 = 1\Omega,\ R5 = 2\Omega$

Answer

In order to find the power loss in R1, we need to know the current flowing through R1. Since it is not in parallel with anything, all of the current flowing through the circuit will flow through R1. To find the current flowing through the circuit, we will need to first find the total equivalent resistance of the circuit.

To do this, we first we need to condense R3 and R4. They are in series, so we can simply add them to get:

$R_{34} = 5\Omega$

Now we can condense R2 and R34. They are in parallel, so we will use the following equation:

$\frac{1}{R_{eq}} = \sum\frac{1}{R} = \frac{1}{3}+\frac{1}{5} = \frac{8}{15}$

$R_{eq}=\frac{15}{8}$

The equivalent circuit now looks like:

Since everything is in parallel, we can simply add everything up:

$R_{total} = R1 + R_{eq}+ R5 = 5\Omega+\frac{15}{8}\Omega+2\Omega$

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

Now that we have the total resistance of the circuit, we can use Ohm's law to find the current:

Rearranging for current, we get:

$I = \frac{V}{R} = \frac{12}{\frac{71}{8}} = \frac{96}{71}=1.35 A$

Now that we know the current flowing through R1, we can use the following equation to find the power loss:

Since we don't know the voltage drop across R1 (although we can calculate it), we can substitute Ohm's law into the equation:

Plugging in our values, we get:

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Question

Consider the given circuit:

If $R1 = 3\Omega$ and $R2 = 2\Omega$ , what is the power loss through R2?

Answer

To calculate the power loss through R2 we need to either calculate the current flowing through it or the voltage drop across it. Calculating the current will be one less step, so we'll use that method.

We need to first calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add their resistance values.

$R_{eq} = R1 + R2 = 3\Omega+2\Omega=5\Omega$

Then we can use Ohm's law to calculate the current through the circuit:

$I=\frac{V}{R}=\frac{12V}{5\Omega} = 2.4A$

Next, use the expression for power:

Substituting Ohm's law for the voltage across the resistor, we get:

$P = I^2R = (2.4A)^2(2\Omega)= 11.5W$

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Question

Consider the given circuit:

What is the total power loss through the circuit if we attach a $4\Omega$ resistor from A to B?

$R1 = 3\Omega$

$R2 = 2\Omega$

Answer

We know the voltage of the circuit, so we simply need the current through the circuit.

The new circuit has two resistors in parallel: R2 and the new one attached. To find the equivalent resistance of these two branches, we use the following expression:

$\frac{1}{R_{eq}}=\sum\frac{1}{R}=\frac{1}{R_2}+\frac{1}{4\Omega}=\frac{1}{2\Omega}+\frac{1}{4\Omega}=\frac{3}{4}$

$R_{eq}=\frac{4}{3}\Omega$

In this new equivalent circuit, everything is in series, so we can simply add up the resistances:

$R_{tot}=R1+R_{eq}=3\Omega+\frac{4}{3}\Omega=4\frac{1}{3}\Omega$

Now we can use Ohm's law to calculate the total current through the circuit:

$V=IR_{tot}$

$I=\frac{V}{R_{tot}}=\frac{12V}{\frac{13}{3}\Omega} = 2.78A$

Now we can use the equation for power:

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Question

Consider the given circuit:

A voltage is applied across points A and B such that the resulting current flows from A to R2 to B. What is the power loss through R2 if the new voltage is ?

$R1 = 3\Omega$

$R2 = 2\Omega$

Answer

To calculate the power loss through R2, we need to know the current flowing through that resistor. Since both voltages cause current to flow in the same direction of R2, we can calculate the current from each and then add them together.

For the original voltage, we need to first calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add them.

$R_{eq} = R1 + R2 =3\Omega+2\Omega= 5\Omega$

Then, we can use Ohm's law to calculate the current through the circuit:

$I=\frac{V}{R}=\frac{12V}{5\Omega} = 2.4A$

Now we can calculate the current caused by the new voltage. There's only one resistor in its path (R2), so we can directly calculate the new current:

$I = \frac{V}{R}=\frac{12V}{2\Omega} = 6A$

Now we can add the two currents together to get the total current through R2:

$I_{tot} = 2.4A + 6A = 8.4A$

Next, we can calculate the power loss of this resistor:

Substituting Ohm's law for voltage:

$P=I^2R=(8.4A)^2(2\Omega) = 141W$

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Question

Consider the circuit:

How much power is dissipated between R2 and R3?

$R1 = 5\Omega,\ R2 = 10\Omega,\ R3 = 4\Omega,\ R4 = 8\Omega$

Answer

We can go about solving this equation by either using Kirchoff's loop rule or Kirchoff's junction rule. Using the loop rule will be much simpler and quicker, so we will go that route.

The loop rule states that through any closed path loop, all voltages must add to zero. For this problem, we will consider two different loops. The first includes only the power supply and R2, and the second includes only the power supply and R3. According to the rule, we know that 12V are lost across both R2 and R3. Therefore, we can write:

$I_2 = \frac{12V}{R2}$

$I_3 = \frac{12V}{R3}$

Since we have expressions for current flowing through each resistor, we can use the expression for power loss:

Adding these two expressions together and substituting in our expressions for current, we get:

$P = \frac{(12V)(12V)}{10\Omega}+\frac{(12V)(12V)}{4\Omega} = 50.4W$

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Question

Consider the circuit:

With this configuration, 20W of power is dissipated throughout the circuit. If the voltage source is doubled, how much power will be dissipated throughout the circuit?

Answer

There are a few ways to solve this problem that involve calculating current and resistance values. However, we will take the simplest route, which doesn't involve either of these calculations. First, we'll start with the power equation written out for both scenarios:

Next we will substitute Ohm's law into each expression:

$I_i = \frac{V_i}{R}$

$I_f = \frac{V_f}{R}$

$P_i = \frac{V_i^2}{R}$

$P_f = \frac{V_f^2}{R}$

Notice how the resistance is not labeled initial or final. If the voltage is doubled, then the current will change, but no changes will have been made to any of the resistors.

We can now divide the two equations for power by each other:

$\frac{P_i}{P_f}= \frac{\frac{V_i^2}{R}}{\frac{V_f^2}{R}} = \frac{V_i^2}{V_f^2}$

Rearranging for final power, we get:

$P_f = P_i \frac{V_f^2}{V_i^2}$

We know that , so we can write:

$P_f = P_i \frac{4V_i^2}{V_i^2}= 4P_i$

This problem emphasizes how advantageous it can be to not substitute values in for your variables until you have your final equation and are ready to solve.

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Question

Consider the circuit:

$R1=R2 = 10\Omega$

How much power is dissipated throughout the entire circuit?

Answer

From the problem statement, we know that $R3 = R4 = 20\Omega$ .

Since we know the value of each resistor, we can calculate an equivalent resistance using the following expression:

$\frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}$

$\frac{1}{R_{eq}}=\sum\frac{1}{10\Omega}+\frac{1}{10\Omega}+\frac{1}{20\Omega}+\frac{1}{20\Omega}=\frac{3}{10}\Omega$

$R_{eq}=\frac{10}{3}\Omega$

Now we can use Ohm's law and the expression for power to solve the problem:

Substituting Ohm's law into the expression for power, we get:

$P = \frac{V^2}{R}= \frac{(12V)^2}{\frac{10}{3}\Omega} = 43.2 W$

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Question

The resistor in Angela's food processor is $30 \Omega$ and has a voltage of across it. If her friend Sam uses it for straight to make his famous coleslaw and must reimburse her $85 \cent$ for the electricity use, how much does the power company charge per kilowatt hour?

Answer

We can use the equation, Voltage = Current(Resistance) to determine:

$120V = (Current)(30\Omega )$ so .

Because Electrical Power = Volts (Current), we can determine that the power Sam used was:

He used the food processor for 3 hours, so he used:

of electricity.

Since he had to pay , the company must charge

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Question

$V_{in}=12V$

$Z_1=10\Omega$

$Z_2=14\Omega$

How much power is dissipated along the circuit above?

Answer

Begin by finding the total resistance of the circuit.

$R_{tot}=Z_1+Z_2=24\Omega$

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

$I=\frac{V_{in}}{R}=\frac{12V}{24\Omega}=0.5A$

The equation for power is:

$P=V_{in}I$

Note that the same value can be found by using an alternative form of the power equation:

$P=\frac{V_{in}^2}{R}=\frac{144V^2}{24\Omega}=6W$

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Question

Given a circuit with one resistor with resistance $R=5 \Omega$ and voltage 100V, determine the power generated in $\frac{J}{s}$ .

Answer

The formula for power given in watts is given by:

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

Plug in known values and solve.

$P=\frac{100^2}{5}=\frac{10000}{5}=2000 \frac{J}{s}$

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Question

What is the power dissipated over a circuit consisting of an overall potential drop of 50V and a current of 0.2A?

Answer

Use the expression for power in a circuit:

Here, I is current and V is voltage, we can multiply the given current and potential drop in the circuit to find the power dissipated.

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Question

Light bulbs give their wattage based on their power output when they are in parallel with a voltage source. For most, that comes from an outlet which typically had a voltage of $\Delta V = 120V$ .

What is the resistance of a 60W lightbulb if it's plugged into a socket with $\Delta V = 120V$ ?

Answer

We can determine the resistance of the 60W lightbulb by using the equation that relates power, voltage, and resistance:

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

Where is resistance, $\Delta V$ is voltage difference in the circuit, and is the power output.

We know that the voltage difference is and that the power output is , so plug in and solve for the resistance:

$R=\frac{\Delta V^2}{P}=\frac{(120)^2V^2}{60J}=240\Omega$

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Question

Suppose that a circuit connected to a battery generates of power across its resistor. What is the resistance of this circuit?

Answer

In this question, we're told that a battery connected circuit is generating a certain amount of power across its resistor, and we're being asked to determine the value of this circuit's resistance.

To start with, recall that the power generated by a circuit is proportional to the applied voltage as well as to the current flowing through the circuit. Written in equation form, we have:

From the above expression, we have values for the power and voltage terms, but we do not have a value for the current. However, we can make use of Ohm's law in combination with the above expression to solve for resistance.

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

By plugging in this expression for current into the above expression for power, we obtain:

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

Next, we can solve for the answer by rearranging and plugging in values.

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

$R=\frac{(5: V)^{2}}{500: W}=\frac{25: V^{2}}{500: W}=0.05: \Omega$

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Question

What happens to current when circuit power is halved and resistance is kept constant?

Answer

To solve this problem, we need an appropriate equation for power that relates current, power, and resistance.

This is given by

, where is power, is current, and is resistance.

We see that current and power are proportional via:

$\sqrt P\propto I$

Since power is changed by a factor of $\frac{1}{2}$ , current changes by

$\sqrt{\frac{1}{2}}$

This can be written alternatively as:

$\sqrt\frac{1}{2}=\frac{1}{\sqrt 2}=\frac{\sqrt 2}{2}$

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Question

A circuit shown below has an electromotive force (emf) of 10V. $R_1=5\Omega$ and $R_2=15\Omega$ . What is the power dissipated by ?

Answer

The power supply (in this case what is providing the emf) will have a power output depending on what is connected to it. A battery or lab power supply is generally designed to put out a constant voltage. The different circuit elements connected will alter the equivalent resistance of the entire circuit, and the power supply will provide a current and power needed to keep the potential (voltage) constant.

There are 3 equations for power dissipation for a resistor in a circuit. They first is:

$P=I * \Delta V$

Where the power dissipated is equal to the product of the current going through the resistor and the voltage drop across it. The second is:

$P=I^{2} R$

The third is:

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

Notice that for each equation we need to know only 2 out of the 3 variables in Ohm's law. Let's chose the second equation with current and resistance. To find the current, notice that there is only a single loop in the circuit since both resistors are connected in series. This means that the total current coming from the power supply is equal to the current going through and the current going through . To find the current coming from the power supply let's find the total equivalent resistance of the circuit.

There are only two resistors connected in series. The equivalent resistance is just the sum of the series resistors:

$R_{eq}=R_{1}+R_{2}$

$R_{eq}=5\Omega+15\Omega=20\Omega$

The current from the power supply is found using Ohm's law:

$I=\frac{V}{R_{eq}}=\frac{10V}{20\Omega}=0.5A$

The power dissipated by the second resistor is then:

$P=I^{2}* R_{2}=(0.5A)^{2}* 15\Omega=3.75W$

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Question

What is the power of a circuit whose current is and equivalent resistance is $0.25\Omega$ ?

Answer

The power in a circuit is determined by the equation $P=I^{2}R$ , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

In our example,

$P=I^{2}R=(30)^{2}\cdot(0.25)=225W$

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Question

What is the power of a circuit whose current is and voltage is ?

Answer

The power in a circuit is determined by the equation $P=I^{2}R$ , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

Since we are given current and voltage, we will also need Ohm's law, , where is the voltage, is the current in the circuit, and is the equivalent resistance of the circuit.

Solving Ohm's law for resistance gives us $R=\frac{V}{I}$ .

Substituting this form of Ohm's law into the power equation gives us

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

The power equation is now in a form that we can solve with the information we are given.

$P=IV=40\cdot9=360W$

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Question

What is the power of a circuit whose voltage is and equivalent resistance is $10\Omega$ ?

Answer

The power in a circuit is determined by the equation $P=I^{2}R$ , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

Since we are given resistance and voltage, we will also need Ohm's law, , where is the voltage, is the current in the circuit, and is the equivalent resistance of the circuit.

Solving Ohm's law for current gives us $I=\frac{V}{R}$ .

Substituting this form of Ohm's law into the power equation gives us

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

The power equation is now in a form that we can solve with the information we are given.

$P=\frac{V^{2}}{R} \right=\frac{3^{2}}{10} \right=0.9W$

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Question

If the current of a circuit is doubled, how is the power of a circuit changed? Assume the resistance of the circuit stays the same.

Answer

The power in a circuit is determined by the equation $P=I^{2}R$ , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

Assuming the resistance stays the same, if the current is doubled, the power will be four times larger.

Expressed mathematically,

If

$P=(2I)^{2}R=4I^{2}R=4P$

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Question

If the voltage of a circuit is doubled, how is the power of a circuit changed? Assume the resistance of the circuit stays the same.

Answer

The power in a circuit is determined by the equation $P=I^{2}R$ , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

To relate power to voltage, we will also need Ohm's law, , where is the voltage, is the current in the circuit, and is the equivalent resistance of the circuit.

Solving Ohm's law for current gives us $I=\frac{V}{R}$ .

Substituting this form of Ohm's law into the power equation gives us

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

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

Assuming the current stays the same, if the voltage is doubled, the power will be four times larger.

Expressed mathematically,

If

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

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