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:

\(\displaystyle R_{34} = 5\Omega\)

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

\(\displaystyle \frac{1}{R_{eq}} = \sum\frac{1}{R} = \frac{1}{3}+\frac{1}{5} = \frac{8}{15}\)

\(\displaystyle R_{eq}=\frac{15}{8}\)

The equivalent circuit now looks like:

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

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

\(\displaystyle R_{total} = 8\frac{7}{8}\Omega\)

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

\(\displaystyle V=IR\)

Rearranging for current, we get:

\(\displaystyle I = \frac{V}{R} = \frac{12}{\frac{71}{8}} = \frac{96}{71}=1.35 A\)

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

\(\displaystyle P = IV\)

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

\(\displaystyle P=I(IR)=I^2R\)

Plugging in our values, we get:

\(\displaystyle P = (1.35)^2(5) = 9.14 W\)

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.

\(\displaystyle R_{eq} = R1 + R2 = 3\Omega+2\Omega=5\Omega\)

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

\(\displaystyle V = IR\)

\(\displaystyle I=\frac{V}{R}=\frac{12V}{5\Omega} = 2.4A\)

Next, use the expression for power:

\(\displaystyle P=IV\)

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

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

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:

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

\(\displaystyle R_{eq}=\frac{4}{3}\Omega\)

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

\(\displaystyle R_{tot}=R1+R_{eq}=3\Omega+\frac{4}{3}\Omega=4\frac{1}{3}\Omega\)

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

\(\displaystyle V=IR_{tot}\)

\(\displaystyle I=\frac{V}{R_{tot}}=\frac{12V}{\frac{13}{3}\Omega} = 2.78A\)

Now we can use the equation for power:

\(\displaystyle P=IV = (2.78A)(12V) = 33.2W\)

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.

\(\displaystyle R_{eq} = R1 + R2 =3\Omega+2\Omega= 5\Omega\)

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

\(\displaystyle V = IR\)

\(\displaystyle I=\frac{V}{R}=\frac{12V}{5\Omega} = 2.4A\)

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:

\(\displaystyle V = IR\)

\(\displaystyle I = \frac{V}{R}=\frac{12V}{2\Omega} = 6A\)

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

\(\displaystyle I_{tot} = 2.4A + 6A = 8.4A\)

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

\(\displaystyle P=IV\)

Substituting Ohm's law for voltage:

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

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:

\(\displaystyle V = IR\)

\(\displaystyle 12V = I_2(R2)\)

\(\displaystyle 12V = I_3(R3)\)

\(\displaystyle I_2 = \frac{12V}{R2}\)

\(\displaystyle I_3 = \frac{12V}{R3}\)

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

\(\displaystyle P = IV\)

\(\displaystyle P_2 = I_2(12V)\)

\(\displaystyle P_3 = I_3(12V)\)

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

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

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:

\(\displaystyle P_i = I_iV_i\)

\(\displaystyle P_f=I_fV_f\)

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

\(\displaystyle V = IR\)

\(\displaystyle I_i = \frac{V_i}{R}\)

\(\displaystyle I_f = \frac{V_f}{R}\)

\(\displaystyle P_i = \frac{V_i^2}{R}\)

\(\displaystyle P_f = \frac{V_f^2}{R}\)

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:

\(\displaystyle \frac{P_i}{P_f}= \frac{\frac{V_i^2}{R}}{\frac{V_f^2}{R}} = \frac{V_i^2}{V_f^2}\)

Rearranging for final power, we get:

\(\displaystyle P_f = P_i \frac{V_f^2}{V_i^2}\)

We know that \(\displaystyle V_f = 2V_i\), so we can write:

\(\displaystyle P_f = P_i \frac{4V_i^2}{V_i^2}= 4P_i\)

\(\displaystyle P_f = 80W\)

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 \(\displaystyle R3 = R4 = 20\Omega\).

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

\(\displaystyle \frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}\)

\(\displaystyle \frac{1}{R_{eq}}=\sum\frac{1}{10\Omega}+\frac{1}{10\Omega}+\frac{1}{20\Omega}+\frac{1}{20\Omega}=\frac{3}{10}\Omega\)

\(\displaystyle R_{eq}=\frac{10}{3}\Omega\)

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

\(\displaystyle V= IR\)

\(\displaystyle P=IV\)

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

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

Begin by finding the total resistance of the circuit.

\(\displaystyle R_{tot}=Z_1+Z_2=24\Omega\)

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

\(\displaystyle I=\frac{V_{in}}{R}=\frac{12V}{24\Omega}=0.5A\)

The equation for power is:

\(\displaystyle P=V_{in}I\)

\(\displaystyle P=12V(0.5A)\)

\(\displaystyle P=6W\)

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

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

The formula for power given in watts is given by:

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

Plug in known values and solve.

\(\displaystyle P=\frac{100^2}{5}=\frac{10000}{5}=2000 \frac{J}{s}\)

Use the expression for power in a circuit:

\(\displaystyle P=IV\)

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.

\(\displaystyle P=0.2*50=10W\)