Using the relationship:

\(\displaystyle \rho =R\frac{A}{l}\)

Here, \(\displaystyle \rho\) is the resistivity, \(\displaystyle R\) is the resistance, \(\displaystyle A\) is cross sectional area, and \(\displaystyle l\) is the length.

The area of a circle is:

\(\displaystyle A=\pi r^2\)

Substitute:

\(\displaystyle \rho =R\frac{\pi r^2}{l}\)

Plug in given values and solve.

\(\displaystyle \rho =.75\Omega \frac{\pi (.008m)^2}{.05m}\)

\(\displaystyle \rho=3.0 *10^{ -3} \Omega \cdot m\)

To calculate the power consumption of the circuit, we need to first reduce it to an equivalent circuit with a single resistor. Since each resistor has the same resistance, this solution will keep resistance calculations as multiples of \(\displaystyle R\) until the circuit is fully reduced.

Start with the two branches in parallel. We can condense R3 and R4, then solve for the total resistance of R2, R3, and R4.

\(\displaystyle R_{34}=R+R=2R\)

\(\displaystyle \frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{R_2}+ \frac{1}{R_{34}}=\frac{1}{R}+\frac{1}{2R} = \frac{3}{2R}\)

\(\displaystyle R_{eq} = \frac{2}{3}R\)

The equivalent circuit now has three resistors in series (R1, Req, and R5), so we can simply add them all up:

\(\displaystyle R_{tot} = R+\frac{2}{3}R+R=\frac{8}{3}R\)

Plug in the value for R:

\(\displaystyle R_{tot}=\frac{8}{3}R=\frac{8}{3}(3\Omega)=8\Omega\)

Now we can use the equation for power:

\(\displaystyle P = IV\)

Substituting in Ohm's law for current, we get:
\(\displaystyle P = \frac{V^2}{R} = \frac{(12V)^2}{8\Omega} = 18W\)

The equation for power is

\(\displaystyle P=VI\)

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 (\(\displaystyle V = IR\)).

To find the total resistance, remember the equations for adding resistors is this:

\(\displaystyle (Series)\ R_{tot}&=R_1+R_2+...+R_n\)

\(\displaystyle (Parallel)\ \frac{1}{R_{tot}}&=\frac{1}{R_1}+\frac{1}{R_2}=...=\frac{1}{R_n}\)

Resistors \(\displaystyle A\) and \(\displaystyle B\) are in series, resistors \(\displaystyle AB\) and \(\displaystyle C\) are in parallel, and resistors \(\displaystyle ABC\) and \(\displaystyle D\) are in series.

\(\displaystyle R_{AB}&=1+3\)

\(\displaystyle R_{AB}= 4\)

\(\displaystyle R_{ABC}&=(\frac{1}{R_{AB}}+\frac{1}{R_C})^{-1}\)

\(\displaystyle R_{ABC}= (\frac{1}{4}+\frac{1}{4})^{-1}\)

\(\displaystyle R_{ABC}= 2\)

\(\displaystyle R_{total}&=R_{ABC}+R_{D}\)

\(\displaystyle R_{total}= 2+2\)

\(\displaystyle R_{total}= 4\)

Now, we can find the current.

\(\displaystyle I&=\frac{V}{R}\)

\(\displaystyle I=\frac{8}{4}\)

\(\displaystyle I= 2\)

Finally, we can find the power.

\(\displaystyle P&=VI\)

\(\displaystyle P=(8)(2)\)

\(\displaystyle P= 16\)

Therefore, the power is 16W (watts).

\(\displaystyle I_{1}=2A\)

\(\displaystyle I_{2}=1A\)

\(\displaystyle R_{1}=4\Omega\)

\(\displaystyle R_{2}=3\Omega\)

\(\displaystyle C_{1}=2F\)

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 \(\displaystyle I3\).

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 \(\displaystyle R1\).

\(\displaystyle I_{1}=I_{2}+I_{3}\)

\(\displaystyle 2=I_{3}+1\)

\(\displaystyle I_{3}=1\)

Now that we know \(\displaystyle I\) and \(\displaystyle R\), we can calculate power across the resistor.

\(\displaystyle P={I^2}R=3W\)

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 \(\displaystyle V\) and each bulb has a resistance of \(\displaystyle R\).

The current going through bulbs A and B is \(\displaystyle \frac{V}{R}\).

However, the current going through bulbs C and D is \(\displaystyle \frac{V}{2R}\).

The current going through bulbs C and D is half as much as the other two, so their brightness will be less.

So, bulbs A and B will be brigher than bulbs C and D.

The equation for power dissipated in a circuit is 

\(\displaystyle P=IV\)

The three resistors are in parallel with each other, so the total resistance is 

\(\displaystyle (R_{tot})^{-1}=\left(\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}\)

\(\displaystyle (R_{tot})^{-1}=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)^{-1}\)

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

Use Ohm's law to find current.

\(\displaystyle I&=\frac{V}{R}\)

\(\displaystyle I=\frac{4}{\frac{8}{7}}\)

\(\displaystyle I= 3.5A\)

Finally, solve for power.

\(\displaystyle P=IV\)

\(\displaystyle P=(4)(3.5)\)

\(\displaystyle P= 14W\)

First, find the total resistance of the circuit.

\(\displaystyle R_1\) and \(\displaystyle R_2\) are in parallel, so we find their equivalent resistance by using the following formula:

\(\displaystyle \frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}\)

\(\displaystyle \frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}\)

\(\displaystyle \frac{3}{6}=\frac{1}{R_{1+2}}}\)

\(\displaystyle R_{1+2}=2\Omega\)

Next, add the series resistors together.

\(\displaystyle R_{1+2}+R_3+R_4=R_{total}\)

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

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

\(\displaystyle V=IR\)

\(\displaystyle (V_1+V_2)=IR\)

\(\displaystyle 9V=I\cdot 8\Omega\)

\(\displaystyle I=\frac{9}{8}A\)

Because it is not in parallel, the total current in the circuit is equal to the current in \(\displaystyle R_3\).

The equation for power is as follows:

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

\(\displaystyle P=\left(\frac{9}{8}A \right )^2\cdot3\Omega\)

\(\displaystyle P=\frac{243}{64}W\)

First, find the total resistance of the circuit.

\(\displaystyle R_1\) and \(\displaystyle R_2\) are in parallel, so we find their equivalent resistance by using the following formula:

\(\displaystyle \frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}\)

\(\displaystyle \frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}\)

\(\displaystyle \frac{3}{6}=\frac{1}{R_{1+2}}}\)

\(\displaystyle R_{1+2}=2\Omega\)

Next, add the series resistors together.

\(\displaystyle R_{1+2}+R_3+R_4=R_{total}\)

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

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

\(\displaystyle V=IR\)

\(\displaystyle (V_1+V_2)=IR\)

\(\displaystyle 9V=I\cdot 8\Omega\)

\(\displaystyle I=\frac{9}{8}A\)

The current through \(\displaystyle R_1\) and \(\displaystyle R_2\) needs to add up to the total current, since they are in parallel.

\(\displaystyle I_1 + I_2 =I_{total}\)

\(\displaystyle I_1 + I_2 =\frac{9}{8}A\)

Also, the voltage drop across them need to be equal, since they are in parallel.

\(\displaystyle V_1=V_2\)

\(\displaystyle I_1R_1=I_2R_2\)

\(\displaystyle I_1\cdot3\Omega=I_2\cdot 6\Omega\)

 Set up a system of equations.

\(\displaystyle I_1\cdot3\Omega=I_2\cdot 6\Omega\)

\(\displaystyle I_1 + I_2 =\frac{9}{8}A\) 

Solve. 

\(\displaystyle I_2=.5I_1\)

\(\displaystyle 3I_2=\frac{9}{8}A\)

\(\displaystyle I_2=\frac{3}{8}A\)

The equation for power is as follows:

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

\(\displaystyle P=\left(\frac{3}{8}A \right )^2\cdot6\Omega\)

\(\displaystyle P=\frac{27}{32}W\)

First, find the total resistance of the circuit.

\(\displaystyle R_1\) and \(\displaystyle R_2\) are in parallel, so we find their equivalent resistance by using the following formula:

\(\displaystyle \frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}\)

\(\displaystyle \frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}\)

\(\displaystyle \frac{3}{6}=\frac{1}{R_{1+2}}}\)

\(\displaystyle R_{1+2}=2\Omega\)

Next, add the series resistors together.

\(\displaystyle R_{1+2}+R_3+R_4=R_{total}\)

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

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

\(\displaystyle V=IR\)

\(\displaystyle (V_1+V_2)=IR\)

\(\displaystyle 9V=I\cdot 8\Omega\)

\(\displaystyle I=\frac{9}{8}A\)

The current through \(\displaystyle R_1\) and \(\displaystyle R_2\) needs to add up to the total current, since they are in parallel.

\(\displaystyle I_1 + I_2 =I_{total}\)

\(\displaystyle I_1 + I_2 =\frac{9}{8}A\)

Also, the voltage drop across them need to be equal, since they are in parallel.

\(\displaystyle V_1=V_2\)

\(\displaystyle I_1R_1=I_2R_2\)

\(\displaystyle I_1\cdot3\Omega=I_2\cdot 6\Omega\)

Set up a system of equations.

\(\displaystyle I_1\cdot3\Omega=I_2\cdot 6\Omega\)

\(\displaystyle I_1 + I_2 =\frac{9}{8}A\)

Solve.

\(\displaystyle I_2=.5I_1\)

\(\displaystyle 1.5I_1=\frac{9}{8}A\)

\(\displaystyle I_1=\frac{3}{4}A\)

The equation for power is as follows:

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

 \(\displaystyle P=\left(\frac{3}{4}A \right )^2\cdot3\Omega\)

 \(\displaystyle P=\frac{27}{16}W\)

\(\displaystyle R_4\) and \(\displaystyle R_5\) are in series with each other, therefore, they will have the same current values. They also have the same resistance values. Thus, they will have the same power dissipated, as \(\displaystyle P=I^2R\).