Series and Parallel - AP Physics 1

Card 0 of 20

Question

A circuit has a resistor with a resistance of $3 \Omega$ followed by three parallel branches, each holding a resistor with a resistance of $5 \Omega$ . What is the total equivalent resitance of the circuit?

Answer

First, we need to calculate the equivalent resistance of the three resistors in parallel. To do this, we will use the following equation:

$\frac{1}{R_{eq}} = \sum\frac{1}{R} = \frac{1}{5}+\frac{1}{5}+\frac{1}{5} = \frac{3}{5}$

$R_{eq} = \frac{5}{3}\Omega$

Now, to get the total equivalent resistance, we can simply add the two remaining values, since they are in series:

$R_{total} = 3\Omega+\frac{5}{3}\Omega = \frac{14}{3}\Omega$

Compare your answer with the correct one above

Question

Consider the given circuit:

A voltage is applied across points A and B so that current flows from A, to R2, to B. What is the value of this voltage if the current through R2 is 4A?

$R1 = 3\Omega$

$R2 = 2\Omega$

Answer

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.

$R_{eq} = R1 + R2 = 3\Omega+2\Omega=5\Omega$

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

$I=\frac{V}{R}=\frac{12V}{5\Omega} = 2.4A$

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

$I_{tot}=I+I_{new}$

$I_{new}=4A$

$I_{new} = 4A-2.4A= 1.6A$

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:

$V=IR = (1.6A)(2\Omega)=3.2V$

Compare your answer with the correct one above

Question

What is the effective resistance of this DC circuit?

Answer

First, let's remind ourselves that the effective resistance of resistors in a series is $R_{eff}=R_{1}+R_{2}+...$ and the effective resistance of resistors in parallel is $R_{eff}=\frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+...}$ .

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

$R_{x}=R_{2}+R_{3}+R_{4}=3\Omega +2\Omega+1\Omega=6\Omega$ (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:

$R_{y}=\frac{1}{\frac{1}{R_{x}}+\frac{1}{R_{6}}}=\frac{1}{\frac{1}{6\Omega }+\frac{1}{2\Omega }}=1.5\Omega$

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:

$R_{eff}=R_{1}+R_{y}+R_{5}=1\Omega+1.5\Omega+4\Omega=6.5\Omega$

Compare your answer with the correct one above

Question

If we have 3 resistors in a series, with resistor 1 having a resistance of $5\Omega$ , resistor 2 having a resistance of $10\Omega$ , and resistor 3 having a resistance of $15\Omega$ , what is the equivalent resistance of the series?

Answer

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

$R_{tot}=R_1+R_2+...+R_n$

$R_{tot}=5\Omega +10\Omega + 15\Omega=30\Omega$

Compare your answer with the correct one above

Question

Two lightbulbs, one graded at and one graded at are connected in series to a battery. Which one will be brighter? What if they are connected in parallel?

Answer

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:

$P=IV=\frac{V^{2}}{R}$

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.

Compare your answer with the correct one above

Question

Consider two circuits: one contains two resistors wired in series, each with a resistance of $20\Omega$ , while the other contains two resistors wired in parallel, one with a resistance of $50\Omega$ and the other with an unknown resistance. The circuits are completely independent, each having its own battery, and each drawing a current of . What must the resistance of the unknown resistor be for the two circuits to have the same total resistance?

Answer

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

$R_{T} = R_{1} + R_{2} +...$

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

$20\Omega + 20\Omega = 40\Omega$

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

$1/R_{T} = 1/R_{1} + 1/R_{2} + ...$

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:

$1/40\Omega = 1/50\Omega + 1/R_{2}$

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

Compare your answer with the correct one above

Question

A circuit is created using a battery and 3 identical resistors, as shown in the figure. Each of the resistors has a resistance of $100\Omega$ . If resistor is removed from the circuit, what will be the effect on the current through resistor ?

Answer

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.

Compare your answer with the correct one above

Question

You are presented with three resistors, each measure $3 \Omega$ . What is the difference between the total resistance of the resistors combined in series, and the total resistance of the resistors combined in parallel?

Answer

Resistors in series:

$R_s=3\Omega+3\Omega+3\Omega$

$R_s=9\Omega$

Resistors in parallel:

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R_p}=\frac{1}{3\Omega}+\frac{1}{3\Omega}+\frac{1}{3\Omega}$

$\frac{1}{R_p}=\frac{3}{3\Omega}$

$R_p=1\Omega$

$R_s-R_p=8\Omega$

Compare your answer with the correct one above

Question

What is the total resistance of three resistors, $100\Omega$ , $10\Omega$ , and $1\Omega$ , in parallel?

Answer

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

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$

Plug in known values and solve.

$\frac{1}{R_{eq}}=\frac{1}{100\Omega}+\frac{1}{10\Omega}+\frac{1}{1\Omega}$

$\frac{1}{R_{eq}}=\frac{111}{100\Omega}$

$R_{eq}=0.9\Omega$

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

Compare your answer with the correct one above

Question

Determine the total charge stored by a circuit with 2 identical parallel-plate capacitors in parallel with area , and a distance of $2 \mu m$ between the parallel plates. Assume the space between the parallel plates is a vacuum. The circuit shows a voltage difference of 10V.

$\epsilon_o= 8.85*10^{-12}\frac{F}{m}$

Answer

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,

$Q=C \Delta V$

For parallel plate capacitors:

$C=\frac{k\epsilon_oA}{d}$

Here, $\epsilon_o= 8.8510^{-12}\frac{F}{m}$ , 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 $d=210^{-6}m$ , which is the distant between the plates.

Plug in known values to solve for capacitance.

$C=\frac{18.8510^{-12}\frac{F}{m}5m^2}{210^{-6}m}= \frac{4.4310^-11Fm}{210^{-6}m}$

$C=2.2110^{-5}F$

Each of the two capacitors has capacitance $C_{capacitor}=2.2110^{-5}F$

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

$C_{circuit}=2C_{capacitance}=22.2110^{-5}F= 4.4210^{-5}F$

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

$Q_{circuit}=C_{circuit}* \Delta V$ , where $Q_{circuit}$ is the total charge stored in the capacitor. Since $\Delta V= 10V$

$Q_{circuit}=4.4210^{-5}F 10V= 4.42*10^{-4}C$

Compare your answer with the correct one above

Question

What is the total current flowing through a system with 2 resistors in parallel with resistances of $2 \Omega$ and $5 \Omega$ , and a battery with voltage difference of 10V?

Answer

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:

$\frac{1}{2\Omega}+\frac{1}{5\Omega}= \frac{1}{R}$

where is the total resistance of the circuit.

$\frac{7}{10\Omega}=\frac{1}{R}$

$\frac{10}{7}\Omega= R$

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

$I=\frac{V}{R}=\frac{10V}{\frac{10}{7}\Omega}=7A$

Compare your answer with the correct one above

Question

Three batteries are connected in series. What is their equivalent voltage?

Answer

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

$V_{s}=\sum V_{n}=V_{1}+V_{2}+...$

In our problem,

$V_{s}=V_{1}+V_{2}+V_{3}=1.5+1.5+1.5=4.5V$

Compare your answer with the correct one above

Question

Three batteries are connected in parallel. What is their equivalent voltage?

Answer

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.

Compare your answer with the correct one above

Question

Three resistors, $R_{1}=15\Omega$ , $R_{2}=10\Omega$ , and $R_{3}=5\Omega$ , arearranged as follows. What is the equivalent resistance of this setup?

Answer

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,

$\frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...$

In our problem,

$\frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}=\frac{2}{30}+\frac{3}{30}$

$\frac{1}{R_{p}}=\frac{5}{30}$

$R_{p}=\frac{30}{5}=6\Omega$

The parallel resistors can now be treated as one resistor with the resistance $R_{p}$ . To find the total resistance, we add the resistance of $R_{p}$ and $R_{3}$ .

$R_{p}+R_{3}=6+5=11\Omega$

Compare your answer with the correct one above

Question

Four resistors, $R_{1}=15\Omega$ , $R_{2}=10\Omega$ , $R_{3}=5\Omega$ and $R_{4}=20\Omega$ , arearranged as follows. What is the equivalent resistance of this setup?

Answer

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,

$\frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...$

In our problem,

$\frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}=\frac{2}{30}+\frac{3}{30}$

$\frac{1}{R_{p}}=\frac{5}{30}$

$R_{p}=\frac{30}{5}=6\Omega$

The parallel resistors can now be treated as one resistor with the resistance $R_{p}$ . To find the total resistance, we add the resistances of $R_{p}$ , $R_{3}$ and $R_{4}$ .

$R_{p}+R_{3}+R_{4}=6+5+20=31\Omega$

Compare your answer with the correct one above

Question

Four configurations of resistors are shown in the figure. Assume all resistors have the same resistance equal to . Rank the different combinations from largest equivalent resistance to smallest equivalent resistance.

Answer

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:

$R_{eq}=\left ( \frac{1}{R}+\frac{1}{R+R}\right )^{-1}=\left ( \frac{1}{R}+\frac{1}{2R}\right )^{-1}=\left ( \frac{3}{2R}\right )^{-1}=\frac{2}{3}R$

(B) These are just two resistors in series,

$R_{eq}=R+R=2R$

(C) This is just one resistor,

$R_{eq}=R$

(D) All three resistors are in parallel,

$R_{eq}=\left ( \frac{1}{R}+\frac{1}{R}+\frac{1}{R}\right )^{-1}=\left ( \frac{3}{R}\right )^{-1}=\frac{R}{3}$

Ranking them we see that the largest to smallest values are

Compare your answer with the correct one above

Question

You have a battery, and wish to arrange a pair of lightbulbs in a way that would make the most amount of light. Both of the lightbulbs have an equal amount of internal resistance.

Would placing the lighbulbs in series or parallel produce the most light?

Answer

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:

$I_{series}=\frac{12V}{2R_{lightbulb}}$

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:

$I_{parallel}=\frac{12V}{R_{lightbulb}}$

From those two equations, it is clear to see that

$I_{series}=\frac{1}{2}I_{parallel}}$ , and that there would be more light if the lightbulbs are placed parallel to one another.

Compare your answer with the correct one above

Question

A voltage is connected in parallel to two resistors (resistor A and B). Resistor A has twice as much resistance as resistor B. What can you conclude about the voltage across the resistors?

Answer

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.

Compare your answer with the correct one above

Question

A circuit is made up of a voltage source with three resistors connected in series. The resistors have a resistance of $4\Omega$ and the current flowing through one of the resistor is . What is the voltage provided by the voltage source?

Answer

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, .

$R_e_q = 4\Omega + 4\Omega + 4\Omega = 12\Omega$

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

$V = I \times R_e_q$

$V = 3A \times 12\Omega = 36V$

Compare your answer with the correct one above

Question

Compared to similar resistors connected in parallel, similar resistors connected in series have current and voltage drop.

Answer

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.

Compare your answer with the correct one above

Tap the card to reveal the answer