Series and Parallel - AP Physics 1

First, we need to calculate the current flow through R2 without the extra voltage attached. We will need to calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add them.

Then, we can use Ohm's law to calculate the current through the circuit:

Now that we have the current, we can calculate the additional current that the new voltage contributes:

There is only one resistor (R2) in the path of the new voltage, so we can calculate what that voltage needs to be to deliver the new current:

First, let's remind ourselves that the effective resistance of resistors in a series is and the effective resistance of resistors in parallel is .

Start this problem by determining the effective resistance of resistors 2, 3, and 4:

(This is because these three resistors are in series.)

Now, the circuit can be simplified to the following:

Next, we will need to determine the effective resistance of resistors and 6:

Again, the circuit can be simplified:

From here, the effective resistance of the DC circuit can be determined by calculating the effective resistance of resistors , 1, and 5:

The total resistance of resistors in a series is the sum of their individual resistances. In this case,

The first step to figuring out this problem is to figure out how resistances of light bulb correlate to the power rating. For a resistor, the power dissipated is:

Thus, there is an inverse relationship between the resistance of the lightbulb and the power rating.

The second step is to take a look at circuit elements in series and in parallel. In series, they share the same current; in parallel they share the same voltage. Thus, for the two lightbulbs in series, the one with the higher resistance (lower wattage) will be brighter, and for a parallel configuration the one with the lower resistance (higher wattage) will be brighter.

The total resistance of a circuit in series can be described by the equation:

The series circuit in ths problem therefore has a resistance of:

The resistance of a circuit wired in parallel has a total resistance of:

We are assuming that the two circuits have the same total resistance, so to find the resistance of the unknown resistor, we set up the following equation:

This, when solved, gives us a resitance of 200 ohms for our unknown element.

Since the resistors and form a parallel network, removing from the circuit increases the resistance of that part of the circuit. Because the new circuit is the series combination of and , the increased resistance leads to lower current in each of these resistors.

Resistors in series:

Resistors in parallel:

The equation for equivalent resistance for multiple resistors in parallel is:

Plug in known values and solve.

Notice that for resistors in parallel, the total resistance is never greater than the resistance of the smallest element.

To determine total charge stored, we need to add up the capacitance of each capacitor(because they are capacitors in parallel) and multiply by the voltage difference. Recall that for capacitors,

For parallel plate capacitors:

Here, , which is the permittivity of empty space, is the dielectric constant, which is since there is only vacuum present, , which is the area of the parallel plates, and , which is the distant between the plates.

Plug in known values to solve for capacitance.

Each of the two capacitors has capacitance

Since the capacitors are in parallel, the total capacitance is the sum of each individual capacitance. The total capacitance in the circuit is given by:

Plug this value into our first equation and solve for the total charge stored.

, where is the total charge stored in the capacitor. Since

First we need to determine the overall resistance of the circuit before we know how much current is flowing through. Since the resistors are in parallel, their resistances will add reciprocally:

where is the total resistance of the circuit.

Now that we've solved for , we know that the current flowing through the circuit can be found using Ohm's law:

The equivalent voltage of batteries connected in series is the sum of the voltage of each battery, or

In our problem,

The equivalent voltage of batteries connected in parallel is equal to the voltage of 1 battery.

In this problem,

Note: Connecting batteries in parallel increases the capacity of the battery.

To find the equivalent resistance of this system, we must first find the equivalent resistance of the resistors in parallel, then evaluate the resistors in parallel.

The parallel resistor equivalence is given by the following equation,

In our problem,

The parallel resistors can now be treated as one resistor with the resistance . To find the total resistance, we add the resistance of and .

To find the equivalent resistance of this system, we must first find the equivalent resistance of the resistors in parallel, then evaluate the resistors in parallel.

The parallel resistor equivalence is given by the following equation,

In our problem,

The parallel resistors can now be treated as one resistor with the resistance . To find the total resistance, we add the resistances of , and .

Let's go through and figure out what the equivalent resistances are.

(A) This is a resistor in parallel with two series resistors. This looks like:

(B) These are just two resistors in series,

(C) This is just one resistor,

(D) All three resistors are in parallel,

Ranking them we see that the largest to smallest values are

The amount of brightness that each light bulb will produce is proportional to the amount of current passing though it. If the lightbulbs were to be placed in series, the total amount of current passing though each bulb would be equal to the voltage of the battery divided by the sum of the resistances of both lightbulbs, given by:

If the lightbulbs were instead to be placed in parallel, the total amount of current passing though would be equal to the voltage across the lightbulb, equal to the voltage of the battery, divided by the internal resistance of the lightbulb:

From those two equations, it is clear to see that

, and that there would be more light if the lightbulbs are placed parallel to one another.

The key hallmark of parallel circuits is that the elements connected in parallel have the same voltage drop across them. It doesn’t matter if the circuit element is a resistor, capacitor, or an inductor; the voltage drop across all elements is the same. This means that the voltage across both resistors A and B is same. On the other hand, circuit elements connected in series have the same current flowing through them; however, they have different voltage drops.

We need to use the principles of circuits and Ohm’s law to solve this question. Recall that circuit elements (in this question resistors) connected in series have the same current flowing through them. The current, therefore, flowing through all three resistors is . To calculate the voltage we need to first calculate an equivalent resistance of the circuit (a single resistor that models the three resistors). Since the resistors are connected in series, we can simply add the resistance of each resistor to get the equivalent resistance, .

Using Ohm’s law we can now calculate the voltage provided.

To solve this question we need to understand the principles of circuits. Circuit elements (such as resistors, capacitors, and inductors) connected in parallel have the same voltage whereas circuit elements connected in series have the same current flowing through them. In parallel circuits, the current hits nodes (regions where two circuit elements branch out and become parallel) and gets split into two different currents, each supplying the individual circuit elements; therefore, the current flowing through parallel connected circuit elements is always less than the total flow of the current. On the other hand, circuit elements in series have the same current flowing through them; therefore, the total current of the circuit flows through each element connected in series. This means that the current flowing through series circuit is higher.

Voltage drop across the parallel circuit elements is equal; therefore, the voltage equals the voltage supplied by the power source. In circuit elements in series, however, the voltage drop is smaller across each element; therefore, resistors in series have smaller voltage drop and larger current than their parallel circuit counterparts.