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:

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:

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

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.