All of the current flowing through the circuit will go through R5, since it is not in parallel with anything. To find the current flowing through the circuit, we will need to first find the total equivalen 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 total current:

\(\displaystyle V=IR\)

Rearranging for current, we get:

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

Kirchoff's current law tells us that the same amount of current entering the junction after R1 must also leave the junction. We also know that the voltage drop across each path of the split is the same.

Consider the following circuit to help visualize things:

\(\displaystyle I_1 = I_2+I_3\)

\(\displaystyle V_2 = V_3+V_4\)

Using Ohm's law to expand the voltages, we get:

\(\displaystyle I_{2}R_{2} = I_{3}R_{3}+I_3R_4 = I_3(R_3+R_4)\)

We now have two equations that we can solve simultaneously:

1. \(\displaystyle I_1 = I_2+I_3\)
2. \(\displaystyle I_{2}R_{2} =I_3(R_3+R_4)\)

Since we are solve for I3, let's rearrange the first equation for I2:

\(\displaystyle I_2 = I_1-I_3\)

Substituting this into the second equation, we get:

\(\displaystyle (I_1-I_3)R_2=I_3(R_3+R_4)\)

Rearranging to solve for I3:

\(\displaystyle I_1R_2-I_3R_2=I_3(R_3+R_4)\)

\(\displaystyle I_1R_2=I_3(R_3+R_4)+I_3R_2\)

\(\displaystyle I_1R_2=I_3(R_2+R_3+R_4)\)

\(\displaystyle I_3 = \frac{I_1R_2}{R_2+R_3+R_4}\)

We have all the values we need, so simply plug in the given values and solve:

\(\displaystyle I_3 = \frac{(1.35)(3)}{(3+4+1)}= \frac{4.05}{8} = 0.51 A\)

Kirchoff's current law tells us that the same amount of current entering the junction after R1 must also leave the junction. We also know that the voltage drop across each path of the split is the same.

Consider the following circuit to help visualize things:

\(\displaystyle I_1 = I_2+I_3\)

\(\displaystyle V_2 = V_3+V_4\)

Using Ohm's law to expand the voltages, we get:

\(\displaystyle I_{2}R_{2} = I_{3}R_{3}+I_3R_4 = I_3(R_3+R_4)\)

We now have two equations that we can solve simultaneously:

1. \(\displaystyle I_1 = I_2+I_3\)
2. \(\displaystyle I_{2}R_{2} =I_3(R_3+R_4)\)

To find V2, we need to calculate I2, so let's rearrange the first equation for I3

\(\displaystyle I_3 = I_1-I_2\)

Substituting this into the second equation, we get:

\(\displaystyle I_2R_2 = (I_1-I_2)(R_3+R_4)\)

Rearranging for I2:

\(\displaystyle I_2R_2=I_1(R_3+R_4)-I_2(R_3+R_4)\)

\(\displaystyle I_2R_2+I_2(R_3+R_4)=I_1(R_3+R_4)\)

\(\displaystyle I_2(R_2+R_3+R_4)=I_1(R_3+R_4)\)

\(\displaystyle I_2 = \frac{I_1(R_3+R_4)}{R_2+R_3+R_4}\)

We have all the values we need, so simply plug in the given values and solve:

\(\displaystyle I_2 = \frac{(1.35)(4+1)}{(3+4+1)}= \frac{6.75}{8} = 0.84 A\)

Now that we know I2, we can use Ohm's law to find the voltage drop across that resistor:

\(\displaystyle V_2 = I_2R_2 = (0.84A)(3\Omega) = 2.53 V\)

In order to calculate the current flowing through the circuit in each scenario, we need to calculate the equivalent resistance in each scenario.

SCENARIO 1: Switch open

When the switch is open, everything is series, so we can simply add all of the resistances up to find the total equivalent resistance. Note that R2 will be excluded.

\(\displaystyle R_{total} = R1+R3+R4+R5 = 5\Omega+4\Omega+1\Omega+2\Omega = 12\Omega\)

Using Ohm's Law, we can now calculate the current flowing through the circuit:

\(\displaystyle V = IR\)

Rearranging for current:

\(\displaystyle I=\frac{V}{R}=\frac{12V}{12\Omega} = 1A\)

SCENARIO 2: Switch closed

The circuit now looks like this:

Since we have a few resistors in parallel, this calcluation will involve a few more steps.

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 series, we can simply add everything up:

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

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

Now using Ohm's law to calculate the current flowing through the circuit:

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

Now, we simply take the different of the currents in the two scenarios:

\(\displaystyle \Delta I = 1.35 A - 1 A = 0.35 A\)

We are asked to compare two different scenarios, each involving the calculation of equivalent resistance, which will use the following formula:

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

Scenario 1: With R3 Present

\(\displaystyle \frac{1}{R_{eq}} = \sum\frac{1}{R}= \frac{1}{5\Omega}+\frac{1}{20\Omega}+\frac{1}{25\Omega}+\frac{1}{10\Omega}\)

\(\displaystyle \frac{1}{R_{eq}} = \frac{39}{100}\Omega\)

\(\displaystyle R_{eq} = 2.56\Omega\)

Now using Ohm's law:

\(\displaystyle V = IR\)

\(\displaystyle I = \frac{V}{R}=\frac{12V}{2.56\Omega}=4.68A\)

Scenario 2: Without R3

\(\displaystyle \frac{1}{R_{eq}} = \sum\frac{1}{R}= \frac{1}{5\Omega}+\frac{1}{20\Omega}+\frac{1}{10\Omega}\)

\(\displaystyle \frac{1}{R_{eq}} = \frac{7}{20}\Omega\)

\(\displaystyle R_{eq} = 2.86\Omega\)

Now using Ohm's law:

\(\displaystyle V = IR\)

\(\displaystyle I = \frac{V}{R}=\frac{12V}{2.86\Omega}=4.20A\)

Calculate the change in current:

\(\displaystyle \Delta I = 4.68A-4.20A = 0.48A\)

Although it is possible to solve this problem by calculating an equivalent resistance, calculating a total current through the circuit, and then using Kirchoff's junction rule to find the current through R1, it is much easier to simply use Kirchoff's Loop rule. What this rule says is that through any closed loop in a circuit, all voltages must add up to zero. Written as an equation:

\(\displaystyle \sum V = 0\)

If we consider the closed loop path consisting of only the power supply and R1, we can use Ohm's law to calculate the current:

\(\displaystyle V = IR\)

\(\displaystyle 12V = I(R1)\)

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

The total current through the circuit is unnecessary to solve this problem. We only need to know the current through the resistor, as well as Kirchoff's loop rule. The loop rule states that for any closed loop through a circuit, the voltages add up to zero.

\(\displaystyle \Delta V = 0\)

For this problem, we will consider the loop that consists solely of the voltage source and R2. From the rule, we know that 12V is lost through the resistor. Using Ohm's law, we can write:

\(\displaystyle V = IR\)

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

Rearranging for R2, we get:

\(\displaystyle R2 = \frac{12V}{I_2} = \frac{12V}{7A}= 1.7\Omega\)

The voltage drop is related to the current and resistance via Ohm's law:

\(\displaystyle V=IR\)

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

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

\(\displaystyle I=2A\)

Since the bulb and the resistor are connected in series, the current is the same in each. Electric current is just the flow of charge through the circuit, so the same amount of charge flows through each in one minute.

Because the TV uses \(\displaystyle 90 kWh\), and it was used for \(\displaystyle .25 hrs\), it must have used 

\(\displaystyle 90 kWh/.25 h = 360 kW\).

\(\displaystyle Power = Voltage \cdot Current\) so:

\(\displaystyle Current = 360 kW/120 V = 3A\)

\(\displaystyle 1A = 1\:coulomb/sec\), and since the TV was used for \(\displaystyle 15 \:min(60) = 900\: sec\)

\(\displaystyle 3 \:coulombs/sec \cdot 900\: sec = 2700 \:coulombs\)