Current and Voltage - AP Physics 1

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:

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

We now have two equations that we can solve simultaneously:

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

Substituting this into the second equation, we get:

Rearranging to solve for I3:

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

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:

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

We now have two equations that we can solve simultaneously:

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

Substituting this into the second equation, we get:

Rearranging for I2:

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

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

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.

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

Rearranging for current:

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:

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

The equivalent circuit now looks like:

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

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

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

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

Scenario 1: With R3 Present

Now using Ohm's law:

Scenario 2: Without R3

Now using Ohm's law:

Calculate the change in current:

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:

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:

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.

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:

Rearranging for R2, we get:

Because the TV uses , and it was used for , it must have used

.

so:

, and since the TV was used for

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.

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

We have to know what SI units are for this problem. SI units are units that cannot be broken down into smaller component parts. Voltage is defined as:

, where is joules and is Coulombs. Coulombs are already in SI units. However, joules is composed of other terms.

Therefore, voltage can be written as

Current and charge are related by the equation where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

Time must first be converted to seconds.

Solving for current,

Current and charge are related by the equation where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

For this problem

Current and charge are related by the equation where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

Time must first be converted to seconds.

Solving for charge,

.

The first step is to find the equivalent resistance of the entire circuit. Note that R2 and R3 are in parallel, and they are in series with R1. Therefore, in order to find the equivalent resistance:

Then, in order to find the total current, use Ohm's Law:

That means that will pass through R1, and will split when it reaches the junction between R2 and R3 based on the ratio of the two resistance:

of current passes through the resistor R3.

To find our answer, we have to know how each of the new additions changes the circuit. The addition of the two resistors in series do not affect the current because current is equivalent for resistors in series. However, since there is a resistor in parallel with equivalent resistance to the sum of the 3 resistors in series, current will split evenly amongst the two pathways, thus leading to a current of flowing through the original resistor.

In order to find the voltage drop across , we will use Ohm's law:

.

However, we first need to solve for the current in the circuit. This will be calculated by applying Ohm's law to the entire circuit, using the total resistance for resistors in series.

Finally, we use this calculated current to find the voltage drop across .

In circuits with resistors, the only thing necessary to determine voltage drop across a resistor is the current through it and the resistance, as given by Ohm's law.

The correct answer is Wire B with and of current.

The formula for voltage, current and resistance is as follows:

Thus, the wire with the lowest ratio of voltage to current will have the least resistance.

To find the current we must first find the equivalent resistance. For resistors in series, the equivalent resistance is

For this problem

Now we use Ohm's law, , to find the current,