To calculate the amount of electrons flowing through this circuit we need to first calculate the current. Use Ohm’s law:

\(\displaystyle V = IR\)

Solve for current:

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

The question gives us the voltage and resistance; therefore, the current flowing through this circuit is

\(\displaystyle I = \frac{10V}{5\Omega } = 2A\)

Recall that current is the amount of electrons flowing through a circuit per unit time.

\(\displaystyle I = \frac{charge\: of\: electrons}{time}\)

Solving for the charge of electrons gives us

\(\displaystyle charge\:of\:electrons = I \times time = 2A \times 5s = 10C\)

Recall that an electron contains \(\displaystyle 1.602\times 10^-19C\); therefore, the number of electrons in this circuit is

\(\displaystyle No.\:of\:electrons = 10C\times \frac{1\: electron}{1.602\times 10^-^19C}\)

\(\displaystyle No. \:of \:electrons = 6.2 \times 10^1^9\)

This means that \(\displaystyle 6.2 \times 10^1^9\) electrons are flowing through this circuit every five seconds.

Using ohm's law, the resistance is determined to be \(\displaystyle \frac{V}{I}\)which is calculated to be \(\displaystyle 2\Omega\). Ohm's law is used again to find the current at \(\displaystyle 30V\) with the same resistance.

Recall Ohm's Law:

\(\displaystyle V=IR,\) where \(\displaystyle V\) is the voltage, \(\displaystyle I\) is the current, and \(\displaystyle R\) is the resistance.

Since the two quantities we are interested in are on the same side of the equation, they are inversely proportional. Hence, if one increases, the other one decreases by the same ratio. Since the current is doubling, the resistance must halve for the circuit to be the same.

To answer this question, we'll need to find an expression for the value of the variable resistor that would make the voltmeter read zero.

First, it's important to realize what situation would result in a reading of zero from the voltmeter. For there to be no reading, that means that there cannot be any voltage difference between the top row of resistors and the bottom row. For this to happen, the voltage drop for the resistors on the left of the voltmeter must be equal, and the same is true for the two resistors to the right of the voltmeter. In other words, both rows of resistors will experience the same voltage decrease as current flows through, thus the difference of voltage drop in the top and bottom row will be identical.

So let's consider the top and bottom resistors on the left side first. In the top left corner, the voltage of the first resistor will be \(\displaystyle I_{top}R_{1}\) from Ohm's law. Moreover, the voltage drop of the bottom left resistor will be \(\displaystyle I_{bottom}R_{2}\). These two voltages will need to be equal to one another in order to have the voltmeter read zero.

\(\displaystyle I_{top}R_{1}=I_{bottom}R_{2}\)

Now let's take a look at the other resistors on the right. The voltage of the third resistor will be \(\displaystyle I_{top}R_{3}\) and the voltage of the variable resistor will be \(\displaystyle I_{bottom}R_{x}\). Just as before, these two resistors will also need to be equal in voltage.

\(\displaystyle I_{bottom}R_{x}=I_{top}R_{3}\)

Now that we have the two expressions shown above, we can isolate the term for the variable resistance in terms of the other three resistors to find our answer.

\(\displaystyle I_{bottom}=\frac{I_{top}R_{1}}{R_{2}}\)

\(\displaystyle R_{x}=\frac{I_{top}R_{3}}{I_{bottom}}\)

\(\displaystyle R_{x}=\frac{I_{top}R_{3}}{(\frac{I_{top}R_{1}}{R_{2}})}\)

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

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

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:

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

Therefore:

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

The equivalent circuit now looks like:

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

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

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

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:

\(\displaystyle \frac{1}{R_{eq}}=\sum\frac{1}{R}=\frac{1}{R_2}+\frac{1}{5\Omega}=\frac{1}{2\Omega}+\frac{1}{5\Omega}=\frac{7}{10}\)

\(\displaystyle R_{eq}=\frac{10}{7}\Omega\)

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

\(\displaystyle R_{tot}=R1+R_{eq}=3\Omega+\frac{10}{7}\Omega=4\frac{3}{7}\Omega\)

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

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

\(\displaystyle I=\frac{V}{R_{tot}}=\frac{12V}{\frac{31}{7}\Omega} = 2.7A\)

We will be working backwards on this problem, using the current to find the resistance. We know the voltage and desired current, so we can calculate the total necessary resistance:

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

\(\displaystyle R_{tot}=\frac{12V}{3A} = 4\Omega\)

Then we can calculate the equivalent resistance of the two resistors that are in parallel (R2 and our unknown):

\(\displaystyle R_{tot}=R1+R_{eq}\)

\(\displaystyle R_{eq}=4\Omega-3\Omega=1\Omega\)

Now we can calculate what the resistance between point A and B:

\(\displaystyle \frac{1}{R_{eq}}=\sum\frac{1}{R}=\frac{1}{R2}+\frac{1}{R_{AB}}\)

Rearranging for the desired resistance:

\(\displaystyle \frac{1}{R_{AB}}=\frac{1}{R_{eq}}-\frac{1}{R2}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2}\)

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

We can use the equation for equivalent resistance of parallel resistors to solve this equation:

\(\displaystyle \frac{1}{R_{eq}}=\sum \frac{1}{R}\)

We know the equivalent resistance, and we know that the resistance of each of the four resistors is equal:

\(\displaystyle \frac{1}{10\Omega} = \frac{4}{R}\)

\(\displaystyle R = 40\Omega\)

Since we know the power loss and voltage of the circuit, we can calculate the equivalent resistance of the circuit using the following equations:

\(\displaystyle V = IR\)

\(\displaystyle P = IV\)

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

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

Rearranging for resistance, we get:

\(\displaystyle R = \frac{V^2}{P}=\frac{(12V)^2}{40W} = 3.6\Omega\)

This is the equivalent resistance of the entire circuit. Now we can calculate R4 using the expression for resistors in parallel:

\(\displaystyle \frac{1}{R_{eq}}=\sum \frac{1}{R}\)

\(\displaystyle \frac{1}{3.6\Omega} = \frac{1}{10\Omega}+\frac{1}{15\Omega}+\frac{1}{20\Omega}+\frac{1}{R4}\)

\(\displaystyle R4 = 16.4\Omega\)

We can use Ohm's law to calculate the equivalent resistance of the circuit:

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

\(\displaystyle R_{eq} = \frac{V}{I}=\frac{12V}{24A}= 0.5\Omega\)

Now we can use the expression for combining parallel resistors to calculate R1:

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

\(\displaystyle \frac{1}{0.5\Omega}=\sum\frac{1}{R1}+\frac{1}{2R1}+\frac{1}{4R1}+\frac{1}{8R1}\)

\(\displaystyle \frac{1}{0.5\Omega}=\frac{15}{8R1}\)

\(\displaystyle R1 = 0.94\Omega\)