Circuits - AP Physics 1

When resistors are in series, they add normally, such as

when in series, they add via their reciprocal

Using these rules, we can first combine all the resistors in series ( and , and ), which can be diagrammed as such:

Using the parallel rule, find to total equivalent resistance.

The relationship between a capacitor's charge and the voltage drop across it is:

Since the voltage drop across both and are the same, we just have to worry about the right part of the circuit. Capacitors are the opposite of resistors when it comes to finding equivalent capacitance, so for capacitors in series the two capacitors on the right will add as such

Plugging into the first equation.

Since the two capacitors are in series they must share the same charge as the equivalent capacitor.

Capacitors add opposite of the way resistors add in a circuit. That is, for capacitors in series they add as such:

Capacitors in parallel add as such:

Use this information to add all the capacitors in series together. The only branch this applies to is the right hand branch.

The equivalent circuit is shown below:

Add the capacitors in parallel.

The first step to solving this problem is to utilize Kirchoff's rules to write a set of equations to find the unknowns. Below two loops are diagrammed, and we assign 3 currents to the circuit. The first and second pass up through batteries 1 and 2 respectively, and the third current passes down through .

These initial conditions give us our first equation:

Now using the loop rule and setting all voltage changes across a loop equal to zero, we get these two equations for the two loops. The loop on the left gives us:

The loop on the right gives us:

Since we're looking to find the current that flows through we will need to solve for . The easiest way to do this would be to take the first equation and replace either or in one of the other two equations. The following solution will replace in the first equation to accomplish this.

Rearrange.

Simplify the second equation.

Lastly, solve for and set the two equations equal to each other.

To find the amperage, first find the combined resistances of the resistors in parallel:

After that, calculate the current using Ohm's Law:

To find the voltage, first find the combined resistances of the resistors in parallel:

Use Ohm's law to find the voltage.

Find the total resistance of the circuit, which can be determined using Ohm's law.

Now, the resistance of the second resistor can be found. Since the two resistors are in parallel, they're related to the total resistance as follows:

Rearrange and solve for

Find the combined resistances for the resistors in parallel:

Combine these two combined series resistors to find the total resistance:

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

Therefore, the total resistance is:

Now, note that since and are in parallel, the voltage drop across them is the same. Use Ohm's law to relate current in terms of voltage and resistance.

Substitute into Ohm's law for the resistance across :

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

Therefore, the total resistance is:

From Ohm's law, we know that is the current traveling through the circuit.

This current will be divided between and , with more current taking the path of lower resistance.

Total voltage drop across :

The current through is given by:

Begin by combining the resistors that are immediately in series:

Now to find the total resistance, combine these two new resistance values, which are in parallel:

To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top and the bottom is the same and equal to the voltage across the circuit, .

The voltage drop across can be found as:

To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top and the bottom is the same and equal to the voltage across the circuit, .

Furthermore, the current that passes through must be the same as the current that passes through .

Therefore, the current that passes through them can be found by rearranging Ohm's law, solving for current.

The quickest way to approach this problem is to realize that the voltage drop across is the same as the voltage drop across the combined resistances of and . Since this parallel combination is the only presence of resistance in the circuit, this voltage drop must be the total voltage of the circuit, .

Therefore, the current across is:

Realize that the voltage drop across the combined resistances of and must be equal to the voltage of the circuit, since the parallel combination is the only presence of resistance in the circuit. This voltage drop must be the total voltage of the circuit, .

The current across and is the same, and is given as:

Begin by finding the resistance of the parallel connection:

The total current is then found using Ohm's law:

The symbol for a capacitor is written as a break in the circuit separated by two parallel lines of equal length as shown below. This loosely resembles the most common type of capacitor, a parallel plate capacitor.

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:

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 parallel, we can simply add everything up:

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

Rearranging for current, we get:

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:

1. 
2. 

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: