Resistors and Ohm's Law: Videos & Practice Problems

Understanding the resistance of a conductor is crucial in electrical circuits. Resistance, denoted as R, opposes the flow of electric current, represented by I, when a voltage V is applied. Ohm's Law, expressed as V = I⁢R, is fundamental for calculating these relationships. Additionally, resistivity ρ quantifies a material's resistance, influenced by its length and cross-sectional area, impacting circuit design and efficiency.

Resistance and Ohm's Law

Understanding the resistance of a conductor is crucial in the study of electricity. When electric charges, such as electrons, flow through a conductor, they encounter atoms within the material. This interaction causes collisions, creating internal friction that resists the flow of current. This phenomenon is quantified by the concept of resistance, denoted by the symbol R, measured in ohms (Ω). The greater the resistance, the less current can flow through the conductor in a given time.

The relationship between voltage (V), current (I), and resistance is encapsulated in Ohm's Law, expressed as:

\( V = I \times R \)

This equation is fundamental in electrical engineering and physics, akin to Newton's second law in mechanics. It indicates that the voltage across a conductor is equal to the product of the current flowing through it and the resistance of the conductor.

To calculate resistance using Ohm's Law, one can rearrange the formula to:

\( R = \frac{V}{I} \)

To find the current, if not directly provided, use the formula:

\( I = \frac{\Delta Q}{\Delta t} \)

where ΔQ is the charge in coulombs and Δt is the time in seconds. For example, if a conductor has a voltage of 10 volts and 6 microcoulombs of charge flows through it in 1.5 seconds, the current can be calculated as:

\( I = \frac{6 \times 10^{-6} \text{ C}}{1.5 \text{ s}} = 4 \times 10^{-6} \text{ A} \)

Substituting the values into the rearranged Ohm's Law gives:

\( R = \frac{10 \text{ V}}{4 \times 10^{-6} \text{ A}} = 2.5 \times 10^{6} \text{ Ω} \)

This calculation illustrates how to determine the resistance of a conductor based on the voltage applied and the current flowing through it. Mastering these concepts and calculations is essential for solving various electrical problems in future studies.

Resistivity & Resistors in Circuits

Understanding the concepts of resistivity and resistance is crucial in the study of electrical circuits. Resistivity, denoted by the Greek letter ρ (rho), is a material property that quantifies how strongly a given material opposes the flow of electric current. The unit of resistivity is ohm-meters (Ω·m). The resistance R of a conductor can be calculated using the formula:

\( R = \frac{\rho L}{A} \)

In this equation, L represents the length of the conductor, and A is its cross-sectional area. The resistivity value is specific to the material; for example, copper, gold, and silver have different resistivity values, which will be provided in problem sets or tests.

When analyzing circuits, resistors are components that provide resistance, and they are typically connected to a power source, such as a battery. In circuit analysis, it is common to assume that connecting wires have negligible resistance, simplifying calculations. However, in practical scenarios, wires do have some resistance, albeit very small.

To illustrate these concepts, consider a wire that is 25.1 meters long with a diameter of 6 millimeters, exhibiting a resistance of 15 milliohms (15 x 10^-3 Ω). To find the resistivity of this wire, we can rearrange the resistance formula:

\( \rho = R \cdot \frac{A}{L} \)

To calculate the area A of the wire, we use the formula for the area of a circle:

\( A = \pi r^2 \)

Here, the radius r is half of the diameter, which is 3 mm or 0.003 m. Substituting the values into the resistivity equation, we find:

\( \rho = (15 \times 10^{-3}) \cdot \frac{\pi (0.003)^2}{25.1} \)

Upon calculating, the resistivity is determined to be approximately \( 1.69 \times 10^{-8} \, \Omega \cdot m \), which is characteristic of copper.

Next, to find the current I flowing through the wire when a voltage V of 23 volts is applied, we can use Ohm's Law:

\( V = I \cdot R \)

Rearranging gives:

\( I = \frac{V}{R} \)

Substituting the known values:

\( I = \frac{23}{15 \times 10^{-3}} \)

This results in a current of approximately \( 1.53 \times 10^{3} \, A \).

These calculations highlight the relationship between resistivity, resistance, and current in electrical circuits, emphasizing the importance of understanding material properties and their implications in practical applications.

A resistor has a current through it of 5 A. If the EMF across the resistor is 10 V, what is the resistance of this resistor?

• A. 2 Ω
• B. 50 Ω
• C. 0.2 Ω
• D. 5 Ω

Answer: (A)

Current Through Unknown Resistor

To determine the current flowing through a cylindrical resistor, we can utilize Ohm's Law, which states that the voltage (V) across a resistor is equal to the product of the current (I) flowing through it and the resistance (R) of the resistor. This relationship can be expressed mathematically as:

$$ V = I \cdot R $$

From this equation, we can rearrange it to solve for current:

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

In this scenario, we are given a voltage (or electromotive force, EMF) of 5 volts. However, to find the current, we first need to calculate the resistance of the cylindrical resistor. The resistance (R) can be determined using the formula:

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

Where:

• $$ \rho $$ is the resistivity of the material (in ohm-meters),
• $$ L $$ is the length of the resistor (in meters), and
• $$ A $$ is the cross-sectional area of the cylinder (in square meters).

For a cylindrical resistor, the cross-sectional area (A) is calculated using the formula for the area of a circle:

$$ A = \pi r^2 $$

Where $$ r $$ is the radius of the cylinder. In this case, we have the following values:

• Resistivity, $$ \rho = 2.2 \times 10^{-7} \, \Omega \cdot m $$
• Length, $$ L = 0.02 \, m $$ (2 centimeters)
• Radius, $$ r = 0.005 \, m $$ (5 millimeters)

Now, we can calculate the cross-sectional area:

$$ A = \pi (0.005)^2 = \pi \cdot 0.000025 \approx 7.85 \times 10^{-5} \, m^2 $$

Next, we substitute the values into the resistance formula:

$$ R = 2.2 \times 10^{-7} \cdot \frac{0.02}{7.85 \times 10^{-5}} $$

Calculating this gives:

$$ R \approx 5.6 \times 10^{-5} \, \Omega $$

Now that we have the resistance, we can find the current using Ohm's Law:

$$ I = \frac{5}{5.6 \times 10^{-5}} $$

This results in:

$$ I \approx 8.9 \times 10^{4} \, A $$

This indicates a very high current flowing through the resistor. Understanding the relationships between voltage, resistance, and current is crucial in electrical engineering and physics, as it allows for the analysis and design of various electrical circuits.

Two resistors are made of the same material, one twice as long as the other. If the current through the shorter resistor is 5 A, what is the current through the longer resistor if they both have the same potential difference and cross-sectional area?

• A. 2.5 Α
• B. 5.0 Α
• C. 10 Α
• D. Not enough information given

Answer: (A)

Here’s what students ask on this topic:

What is Ohm's Law and how is it used in electrical circuits?

Ohm's Law is a fundamental principle in electrical circuits, expressed as $V=IR$. It states that the voltage (V) across a conductor is directly proportional to the current (I) flowing through it, with the proportionality constant being the resistance (R). This law is crucial for calculating the relationships between voltage, current, and resistance in circuits. For example, if you know the voltage and resistance, you can determine the current using $I=\frac{V}{R}$. Understanding Ohm's Law helps in designing and analyzing electrical circuits effectively.

How do you calculate the resistance of a conductor using its resistivity?

The resistance (R) of a conductor can be calculated using its resistivity (ρ) with the formula $R=\frac{\rho L}{A}$, where L is the length of the conductor and A is its cross-sectional area. Resistivity is a material-specific property that quantifies how strongly a material opposes the flow of electric current. For example, if you have a copper wire with a length of 2 meters and a cross-sectional area of 0.0001 square meters, and the resistivity of copper is $1.68\times 10⁻8 \Omega m$, you can calculate the resistance as $0.336\Omega$.

What is the difference between resistance and resistivity?

Resistance (R) and resistivity (ρ) are related but distinct concepts in electrical circuits. Resistance is a measure of how much a specific object opposes the flow of electric current, and it depends on the object's dimensions and material. It is given by the formula $R=\frac{\rho L}{A}$, where L is the length and A is the cross-sectional area. Resistivity, on the other hand, is an intrinsic property of a material that quantifies how strongly the material itself opposes current flow, regardless of its shape or size. It is measured in ohm-meters (Ω·m). While resistivity is a material constant, resistance varies with the dimensions of the conductor.

How do you calculate current using Ohm's Law?

To calculate the current (I) using Ohm's Law, you can rearrange the formula $V=IR$ to solve for I: $I=\frac{V}{R}$. This means that the current is equal to the voltage (V) divided by the resistance (R). For example, if you have a circuit with a voltage of 12 volts and a resistance of 4 ohms, the current would be $3A$ (amperes). This calculation is essential for understanding how much current flows through a circuit given a specific voltage and resistance.

What are the units of resistance and resistivity?

The unit of resistance (R) is the ohm (Ω), which quantifies how much a conductor opposes the flow of electric current. One ohm is defined as the resistance between two points of a conductor when a constant potential difference of one volt applied between these points produces a current of one ampere. The unit of resistivity (ρ) is the ohm-meter (Ω·m). Resistivity measures how strongly a material opposes the flow of electric current per unit length and cross-sectional area. These units are fundamental in electrical engineering and physics for analyzing and designing circuits.