Resistors - Physics

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Question

Ten resistors, each with $300\Omega$ resistance, are set up in series. What is their equivalent resistance?

Answer

For resistors aligned in series, the equivalent resistance is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$

Since all the resistors in this problem are equal, we can simplify with multiplication.

$R_{eq}=10 (300\Omega)$

$R_{eq}=3000\Omega$

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Question

Ten resistors, each with $300\Omega$ resistance, are set up in parallel. What is the equivalent resistance?

Answer

For resistors in parallel, the equivalent resistance can be found by summing the reciprocals of the individual resistances, then taking the reciprocal of the resultant sum.

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

Since all the resistors in this problem are equal, we can simplify by multiplying.

$\frac{1}{R_{eq}}=10( \frac{1}{300\Omega})$

$\frac{1}{R_{eq}}=\frac{10}{300\Omega}$

$\frac{1}{R_{eq}}=\frac{1}{30\Omega}$

$R_{eq}=30\Omega$

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Question

There are three resistors in parallel in a circuit with resistances of $10\Omega$ , $20\Omega$ , and $30\Omega$ .

What is the equivalent resistance?

Answer

The equation for resistors in parallel is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

We are given the values of the resistors. Using this formula, we can solve for the equivalent resistance.

Plug in the given values and solve.

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

$\frac{1}{R_{eq}}=\frac{1}{10\Omega}+\frac{1}{20\Omega}+\frac{1}{30\Omega}...$

$\frac{1}{R_{eq}}=\frac{11}{60\Omega}$

$R_{eq}=\frac{60\Omega}{11}$

$R_{eq}=5.45\Omega$

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Question

Twelve resistors of equal resistance are set up in series. If the total resistance is $600\Omega$ , what is the resistance of each resistor?

Answer

The total resistance of resistors in series is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$

In this case, we know the total resistance and the number of resistors, and we are told that all resistors are of equal strength. That means we can simplify this problem using multiplication.

$R_{eq}=12R_$

We can use the total resistance to solve for the individual value.

$600\Omega=12R$

$\frac{600\Omega}{12}=R$

$50\Omega=R$

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Question

Three resistors are in a series. They have $10\Omega$ , $18.3\Omega$ , and $21.22\Omega$ of resistance respectively. What is the total resistance?

Answer

When working in a series, the total resistance is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$

Use the given values for each individual resistor to solve for the total resistance.

$R_{eq}=10\Omega+18.3\Omega+21.22\Omega$

$R_{eq}=49.52\Omega$

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Question

Six resistors are arranged in a series. Their resistances are $32.4\Omega$ , $8.7\Omega$ , $0.77\Omega$ , $65.2\Omega$ , $10\Omega$ , and $26.33\Omega$ . What is their total resistance?

Answer

When working in a series, the total resistance is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$

Use the given values for each individual resistor to solve for the total resistance.

$R_{eq}=32.4\Omega + 8.7\Omega +0.77\Omega + 65.2\Omega + 10\Omega + 26.33\Omega$

$R_{eq}=143.4\Omega$

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Question

Three resistors are in a parallel circuit. They have resistances of $10\Omega$ , $18.3\Omega$ , and $21.22\Omega$ , respectively. What is their total resistance?

Answer

The formula for resistors in parallel is:

$\frac{1}{R_{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

Using the given individual resistances, we can find the total resistance of the circuit.

$\frac{1}{R_{eq}}}=\frac{1}{10\Omega}+\frac{1}{18.3\Omega}+\frac{1}{21.22\Omega}$

$\frac{1}{R_{eq}}}=0.2017\Omega$

$R_{eq}=4.96\Omega$

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Question

Two resistors in series have $1.8\Omega$ and $32.4\Omega$ resistance, respectively. What is the total resistance?

Answer

For resistors in a series, total resistance is equal to the sum of each individual resistance.

$R_{eq}=R_1+R_2+R_3...$

We can use the individual resistances from the question to solve for the total resistance.

$R_{eq}=1.8\Omega+32.4\Omega$

$R_{eq}=34.2\Omega$

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Question

What is the total resistance of a series circuit with resistors of $2\Omega$ , $0.55\Omega$ , and $4\Omega$ ?

Answer

For a series circuit, the formula for total resistance is:

$R_{eq}=R_1+R_1+R_3...$

We are given the values of each resistance, allowing us to sum them to find the total resistance in the circuit.

$R_{eq}=2\Omega+0.55\Omega+4\Omega$

$R_{eq}=6.55\Omega$

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Question

What is the total resistance of a parallel circuit with resistors of $2\Omega$ , $0.55\Omega$ , and $4\Omega$ ?

Answer

The formula for resistance in parallel is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

We are given the values for each individual resistor, allowing us to solve for the total resistance.

$\frac{1}{R_{eq}}=\frac{1}{2\Omega}+\frac{1}{0.55\Omega}+\frac{1}{4\Omega}$

$\frac{1}{R_{eq}}=\frac{113}{44}\Omega$

$R_{eq}=\frac{44}{113}\Omega=0.389\Omega$

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Question

Calculate the resistance of a copper wire with cross-sectional area of $\small 0.501cm^2$ and length of $\small 2m$ .

$\small rho_{Cu}=1.68*10^{-3}\Omega m$

Answer

The resistance of a wire is given by the following equation:

$R=\rho\frac{L}{A}$

We are given the resistivity ( $\small \rho$ ), cross-sectional area, and length. Using these values, we can solve for the resistance.

First, convert the cross-sectional area to square-meters.

$0.501cm^2\frac{1m^2}{10000cm^2}=5.0110^{-5}m^2$

Use the resistance equation to solve.

$R=(1.6810^{-3}\Omega m)\frac{2m}{5.0110^{-5}m^2}$

$R=67.1\Omega$

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Question

What is the total resistance of a $14\Omega$ , $2.2\Omega$ , and $13\Omega$ series circuit?

Answer

The formula for resistors in series is $R_{eq}=R_1+R_2+R_3...$ .

Plug in our given values.

$R_{eq}=R_1+R_2+R_3...$

$R_{eq}=14\Omega+2.2\Omega+13\Omega$

$R_{eq}=29.2\Omega$

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Question

Two aluminum wires have the same resistance. If one has twice the length of the other, what is the ratio of the diameter of the longer wire to the diameter of the shorter wire?

Answer

Let’s start with the resistivity equation

$R = \rho \frac{L}{A}$

The area of a wire is the area of a circle. So let’s substitute that into our equation

$A = \pi r^2$

$A = \pi \frac{d^2}{4}$

$R = \rho \frac{L}{\pi \frac{d^2}{4}}$

This can be simplified to

$R = 4\rho \frac{L}{\pi d^2}$

Since we know that both resistors have the same resistivity and the same resistance, we can set these equations equal to each other.

$4\rho \frac{L_1}{\pi d_1^2} = 4\rho \frac{L_2}{\pi d_2^2}$

Many things fallout which leaves us with

$\frac{L_1}{d_1^2} = \frac{L_2}{d_2^2}$

We know that the second wire is twice the length as the first

So we can substitute this into our equation

$\frac{L_1}{d_1^2} = \frac{2L_1}{d_2^2}$

The length of the wire drops out of the equation

$\frac{1}{d_1^2} = \frac{2}{d_2^2}$

Now we can solve for the diameter of the longer wire.

Take the square root of both sides

$d_2 = \sqrt{2} d_1$

Therefore the ratio of the long to the short wire is

$\frac{\sqrt{2} d_1}{d_1}$

Or

$\sqrt{2}$

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Question

What is the diameter of a length of tungsten wire whose resistance is ohms?

Answer

We will use the resistivity equation to solve for this. We know

Length =

Resistance =

ρ of Tungsten = $5.6x10^-8 \Omega m$

The equation for resistivity is

$R = \rho \frac{L}{A}$

We can rearrange this equation to solve for

$A =\rho \frac{R}{L}$

In this case the area of the wire is the area of a circle which is equal to $\pi r^2$

We can rearrange this to get the radius by itself

$r = \sqrt{\frac{\rho R}{L \pi}}$

$r = \sqrt{\frac{(5.6x10^-8)(0.32)}{(1)(\pi)}}$

To find the diameter we need to multiply this value by 2.

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Question

Ten resistors, each with $300\Omega$ resistance, are set up in series. What is their equivalent resistance?

Answer

For resistors aligned in series, the equivalent resistance is the sum of the individual resistances.

$R_{eq}=R1+R2+R3...$

Since all the resistors in this problem are equal, we can simplify with multiplication.

$R_{eq}=10(300\Omega)$

$R_{eq}=3000\Omega)$

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Question

What is the total resistance of a parallel circuit with resistors of $2\Omega$ , $0.55\Omega$ , and $4\Omega$ ?

Answer

The formula for resistance in parallel is:

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

We are given the values for each individual resistor, allowing us to solve for the total resistance.

$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{0.55} + {1}{4}$

$\frac{1}{R_{eq}} = \frac{113}{44}$

$R_{eq}= \frac{44}{113} = 0.389$

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Question

The resistivity of most common metals __________________ .

Answer

At higher temperature, the atoms are moving more rapidly and are arranged in a less orderly way. Therefore it is expected that these fast moving atoms are more likely to interfere with the flow of electrons. If there is more interference in the flow of electrons, then there is a higher resistivity.

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Question

An electric circuit is set up in series with five resistors. If the resistors remain the same, but the circuit is now set up with the resistors in parallel, how would this affect the total resistance?

Answer

Think of resistors as doors, preventing the flow of people (electrons). Imagine the following scenarios: a large group of people are in a room and all try to leave at once. If the five resistors are in series, that's like having all of these people trying to go through all five doors before they can leave. In a circuit, all the electrons in the current mast pass through every resistor.

If the resistors are in parallel, it's like having five separate doors from the room. All of a sudden, the group can leave MUCH faster, encountering less resistance to their flow out of the room. The path of the electrons can split, allowing each particle to pass through only one resistor.

From a formula perspective, the resistors in series are simply summed together to find the equivalent resistance.

$R_{eq}=R_1+R_2+R_3 ...$

In parallel, however, the reciprocals are summed to find the reciprocal equivalent resistance.

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

Adding whole numbers will always give you a much greater result than adding fractions. For the exact same set of resistors, arrangement in series will have a greater total resistance than arrangement in parallel.

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Question

When current in a circuit crosses a resistor, energy is lost. What form does this lost energy most commonly take?

Answer

In basic resistors, energy lost due to resistance is converted into heat. In some cases, other conversions also take place (such as generation of light in a lightbulb), but heat is still dissipated along with any alternative conversations. Lightbulbs, batteries, and other types of resistors will become hot as current passes through them.

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Question

You have a long, diameter copper wire that has an electric current running through it. Which of the following would decrease the wire's overall resistivity?

Answer

The formula for resistance is:

$R=\rho LA$

Above, is length, is the cross-sectional area of the wire, and $\rho$ is the resistivity of the material, and is a property of that material. The resistivity is constant for a given material, and thus cannot be changed by altering the dimensions of the wire.

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