Physics - Resistivity | Practice Hub

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Which Resistors are connected in parallel?

All answer choices are correct

$R_{1}$ and $R_{2}$

$R_{2}$ and $R_{3}$

$R_{1}$ and $R_{3}$

None of the answer choices are correct

Explanation

Resistors are in parallel when the electric current passes through two or more branches or connected parts at the same time before it combines again. After the current leaves the battery it travels to $R_{2}$ . Then the current splits and travels between the other two resistors before the current coming back together and connecting back to the battery.

The resistivity of most common metals __________________ .

remains constant over wide temperature ranges

increases as the temperature increases

decreases as the temperature increases

varies randomly as the temperature increases

Explanation

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.

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

Explanation

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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Which resistor has the greatest current going through it?

Assume all the resistors are equal.

$R_{1}$

$R_{1}$ and $R_{2}$

$R_{3}$ and $R_{4}$

$R_{5}$

Explanation

As we examine the circuit we first come across the first two resistors. These resistors are in parallel. In parallel circuits, the current splits to go down each branch. In this case, the resistors are of equal value meaning that the current is split evenly between the two.

We then come across another parallel branch with two resistors on one side and a single resistor on the other side. The two resistors in series add up to create more resistance on the top branch. The single resistor on its own has less resistance.

Since current always chooses the path of least resistance, more current will flow through the single resistor, than through the branch with two resistors in series.

Therefore, would have the greatest resistance flowing through it.

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

$0.153\Omega$

$0.389\Omega$

$0.458\Omega$

$2.57\Omega$

$6.55\Omega$

Explanation

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.

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

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

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

How does adding resistors in parallel affect the overall current of the circuit?

The current increases

The current decreases

The current stays the same

Explanation

Adding resistors in parallel decrease the overall equivalent resistance as they are added using the equation

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

Since the overall equivalent resistance is decreased, if the voltage is constant the overall current would increase.

Therefore

$I = \frac{V}{R}$

This shows the inverse relationship between the two values.

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?

$\frac{1}{\sqrt{2}}$

$\sqrt{2}$

$2\sqrt{2}$

Explanation

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

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

$\sqrt{2}$

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

The energy is converted into motion

The energy is converted into light

The energy is converted into heat

The energy is converted into sound

The energy is not converted; it simply disappears

Explanation

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.

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

$30\Omega$

$3,000\Omega$

$300\Omega$

$30,000\Omega$

$3\Omega$

Explanation

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.

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

$R_{eq}=3000\Omega)$

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?

Increasing the length of the wire

Increasing the diameter of the wire

Decreasing the diameter of the wire

Decreasing the length of the wire

None of these

Explanation

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.

Resistivity

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