Circuit Power - AP Physics 1

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.

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

Next, use the expression for power:

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

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:

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

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

Now we can use the equation for power:

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.

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

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:

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

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

Substituting Ohm's law for voltage:

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:

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:

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:

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:

Rearranging for final power, we get:

We know that , so we can write:

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.

From the problem statement, we know that .

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

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:

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

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

Begin by finding the total resistance of the circuit.

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

The equation for power is:

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

The formula for power given in watts is given by:

Plug in known values and solve.

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.

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

Where is resistance, 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:

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.

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

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

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:

Since power is changed by a factor of , current changes by

This can be written alternatively as:

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:

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:

The third is:

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:

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

The power dissipated by the second resistor is then:

The power in a circuit is determined by the equation , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

In our example,

The power in a circuit is determined by the equation , 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 .

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

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

The power in a circuit is determined by the equation , 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 .

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

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

The power in a circuit is determined by the equation , 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

The power in a circuit is determined by the equation , 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 .

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

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

Expressed mathematically,

If