AP Physics 1 - Series and Parallel

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

$0.9\Omega$

$1.1\Omega$

$9.0\Omega$

$11.1\Omega$

$111\Omega$

Explanation

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.

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?

$30\Omega$

$10\Omega$

$20\Omega$

$40\Omega$

Explanation

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

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

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

Resistors

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.

Explanation

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,

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

$R_{eq}=R$

(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

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? Screen shot 2015 09 08 at 12.11.44 pm

$11\Omega$

$6\Omega$

$10\Omega$

$5\Omega$

Explanation

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 $R_{3}$ .

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

What is the effective resistance of this DC circuit?

$6.5\Omega$

$13\Omega$

$5.7\Omega$

None of the other answers

$2\Omega$

Explanation

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:

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

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?

$I=\frac{10}{7}A$

Explanation

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:

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

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?

Explanation

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.

$V = I \times R_e_q$

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

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?

Series:

Parallel:

There's not enough information to complete this problem

Series:

Parallel:

Series:

Parallel:

Series:

Parallel:

Series:

Parallel:

Explanation

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.

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.

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

$4.42*10^{-4}C$

$1.11*10^{-4}C$

$2.21*10^{-5}C$

$6.44*10^{-5}C$

Explanation

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, $\epsilon_o= 8.85*10^{-12}\frac{F}{m}$ , 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 $d=2*10^{-6}m$ , which is the distant between the plates.

Plug in known values to solve for capacitance.

$C=\frac{1*8.85*10^{-12}\frac{F}{m}*5m^2}{2*10^{-6}m}= \frac{4.43*10^-11Fm}{2*10^{-6}m}$

$C=2.21*10^{-5}F$

Each of the two capacitors has capacitance $C_{capacitor}=2.21*10^{-5}F$

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:

$C_{circuit}=2*C_{capacitance}=2*2.21*10^{-5}F= 4.42*10^{-5}F$

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 $\Delta V= 10V$

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

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?

$200\Omega$

$40\Omega$

$400\Omega$

$160\Omega$

$50\Omega$

Explanation

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.

Series and Parallel

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