19.6 Resistance
19.6.1 Relationship between p.d. and Current through a Conducting Material
1. Relationship between V and I
(a) Suppose a p.d. V is applied across a conducting material so that it produces a current I in the material. We would like to know how a change in V would affect I. In other words, we would like to establish a relationship between V and I.
 |
| Figure 19.7 Simple circuit to study electrical conductivity. A voltmeter (V) measures voltage across the conductor, an ammeter (A) measures current, and a rheostat adjusts the current to analyze the V–I relationship. |
(b) To study the electrical conductivity of materials, a simple circuit can be set up as shown in Figure 19.7. By varying the voltage (V) across a conductor and measuring the current (I), we can record the values in a table:
| V (V) | I (A) |
|---|---|
| --- | --- |
| --- | --- |
| --- | --- |
Plot a graph of I against V (or V against I).
(c) Each graph given in Figure 19.8 shows how I thorough the material varies with V.
We notice that
(a) for some materials (e.g. copper, CuSO4 solution) we have the relationship
$V\propto I$
(b) for the rest of the materials shown, V is not proportional to I.
Current-Voltage (I–V) Graphs for Various Materials
 |
| Figure 19.8 Graph showing the relationship between current (I) and voltage (V). Ohmic materials have a linear I–V curve, while non-ohmic materials show a non-linear behavior. |
- (a) Copper: Linear graph, obeys Ohm's law.
- (b) CuSO4 solution: Linear, Ohmic behavior.
- (c) Dilute H2SO4 acid: Linear, Ohmic.
- (d) Carbon: Non-linear, V not proportional to I.
- (e) Thermistor: Non-linear, resistance decreases with temperature.
- (f) Gas at low pressure: Non-linear I–V curve.
- (g) Vacuum tube diode: Non-linear, exponential behavior.
- (h) Semiconductor diode: Non-linear, exponential rise.
19.6.2 Ohm's Law
1. Definition of the law
(a) A conducting material which obeys the relationship
$V\propto I$
at constant temperature is said to obey Ohm's law.
(b) The graph in Figure 19.8(a) shows that copper obeys Ohm's law.
(c) A material which obeys Ohm's law is known as an ohmic conductor. Most metallic conductors at constant temperature are ohmic conductors.
2. Equivalence between Ohm's law and the relationship $J=\sigma E$
A p.d. V is applied across a length L of a conductor of uniform cross sectional area A. A constant current I flows in the conductor, producing a current dencity J given by
$J=\frac{I}{A}$
The electric field, assumed to be uniform, is set up within the conductor. Its strength E is given by
$E=\frac{V}{L}$
$J=\sigma E$
$\frac{I}{A}=\sigma \frac{V}{L}$
$V=\left(\frac{L}{\sigma A}\right)I$
At constant temperature, $\sigma$, L and A are constant. Hence, we have
$V\propto I$
This relationship is Ohm's law. Hence a conducting material which obeys Ohm's law will also obey the relationship $J\propto E$.
19.6.3 Resistance
1. Definition
The ratio of the p.d. V applied across a material to the current I that flows through the material is defined as the resistance R of the material.
2. Defining equation
$R=\frac{V}{I}$
3. Unit
The unit of R is $V \ A^{-1}$ or ohm ($\Omega$)
EXAMPLE 19.8
Determine the p.d. across a $10 \ \Omega$ resistor if 1200 C of charge flows pass a cross section of the resistor pe minute.
Answer
$V=IR=\left(\frac{q}{t}R\right)=\left(\frac{1200}{60}(10)\right)=200 \ V$
EXAMPLE 19.9 Flashlight Bulb Resistance
A small flashlight bulb draws 300 mA from its 1.5 V battery.
(a) What is the resistance of the bulb?
(b) If the battery becomes weak and the voltage drops to 1.2 V, how would the current change? Assume the bulb is approximately ohmic.
Approach
We apply Ohm’s law to the bulb, where the voltage applied across it is the battery voltage.
Jawab
(a) Convert 300 mA to 0.30 A and use Ohm’s law:
$R = \frac{V}{I} = \frac{1.5 \ V}{0.30} \ A = 5.0 \ \Omega$
(b) If the resistance remains the same, the current is:
$I = \frac{V}{R} = \frac{1.2 \ V}{5.0 \ \Omega} = 0.24 \ A = 240 \ mA$
This represents a decrease of 60 mA.
With the smaller current in part (b), the bulb filament’s temperature would be lower and the bulb less bright. Also, resistance does depend on temperature.
EXAMPLE 19.10
A bird stands on a dc electric transmission line carrying 4100 A (Figure 19.10. The line $2.5 \times 10^{-5} \Omega$ has resistance per meter, and the bird’s feet are 4.0 cm apart. What is the potential difference between the bird’s feet?
Answer
Potential difference between a bird’s feet on a high-voltage line
Given:
$I = 4100 \ A$
$R/m = 2.5 \times 10^{-5} \ \Omega/m$
$L = 0.04 \ m$
Step 1: Understanding the concept
A potential difference (voltage) appears when current flows through a resistance.
Even though the transmission line is very long, we only consider the small segment between the bird’s feet (0.04 m).
Step 2: Calculate the resistance of the wire segment
Use:
$R=R_{per \ m} \times L$
$R=(2.5 \times 10^{-5}) (0.04)=1.0 \times 10^{-6} \ \Omega$
Step 3: Calculate the potential difference
Apply Ohm’s Law:
$V=IR=(4100)(1.0 \times 10^{-6})=0.0041 \ V=4.1 \ mV$
This value is very small (millivolts), meaning there is almost no potential difference between the bird’s feet.
EXAMPLE 19.11
A hair dryer draws 13.5 A when plugged into a 120-V line.
(a) What is its resistance?
(b) How much charge passes through it in 15 min? (Assume direct current.)
Answer
Given:
Current: I = 13.5 A
Voltage: V = 120 V
(a) Resistance:
Step 1: Use Ohm’s Law
Rearrange the formula to solve for resistance:
$R = \frac{V}{I}$
Step 2: Substitute the values
$r=\frac{120}{13.5} ≈ 8.89 \Omega$
The resistance tells us how much the hair dryer opposes the flow of electric current. A value of about 8.89 Ω indicates moderate resistance typical of heating devices.
(b) Charge:
Step 1: Use the definition of current
$I=\frac{Q}{t}$
Rearrange to solve for charge:
$Q=It$
Step 2: Substitute the values
$Q = 13.5 \times 900 = 12150 \ C \cong 1.22 \times 10^4 \ C$
Charge represents the total amount of electric charge flowing through the hair dryer in 15 minutes. Since the current is quite large and the time is long, the total charge is also large.
EXAMPLE 19.12
A 4.5-V battery is connected to a bulb whose resistance is 1.3 Ω. How many electrons leave the battery per minute?
Answer
Given:
Voltage: $V=4.5 \ V$
Resistance: $I = 1.3 \ A$
Time: $t = 1 \ minute = 60 \ s$
Charge of one electron: $e=1.6 \times 10^{-19} \ C$
Step 1: Find the current
Use Ohm’s Law:
$V=IR$
Rearrange to solve for current:
$I=\frac{V}{R}$
Substitute the values:
$I = \frac{4.5}{10} = 0.45 \ A$
The current tells us how much charge flows per second through the circuit. A current of 0.45 A means 0.45 coulombs of charge pass a point every second.
Step 2: Find the total charge
Use the definition of current:
$Q=I \times t$
Substitute the values:
$Q = (0.45)(60) = 27 \ C$
This means that 27 coulombs of charge flow through the circuit in one minute.
Step 3: Find the number of electrons
Use:
$n=\frac{Q}{e}$
Substitute the values:
$n=\frac{27}{1.6 \times 10^{-19}} \cong 1.69 \times 10^{20} \ electrons$
Each electron carries a very small charge, so a large number of electrons is needed to make up 27 coulombs.
EXAMPLE 19.13
An electric device draws 5.60 A at 240 V.
(a) If the voltage drops by 15%, what will be the current, assuming nothing else changes?
(b) If the resistance of the device were reduced by 15%, what current would be drawn at 240 V?
Answer
Given:
Initial current: $I = 5.6 \ A$
Initial voltage: $V=240 \ V$
Step 1: Find the resistance of the device
Use Ohm’s Law:
$V=IR$
Rearrange to solve for resistance:
$R=\frac{V}{I}$
Substitute the values:
$R = \frac{240}{5.60} ≈ 42.857 \ \Omega$
The resistance is assumed to remain constant in part (a), so we first determine its value using the original conditions.
(a) Current when voltage drops by 15%
Step 2: Find the new voltage
$V_{new}=0.85 \times 240=204 \ V$
A 15% drop means the voltage becomes 85% of its original value.
Step 3: Find the new current
$I=\frac{V_{new}}{R}$
$I = \frac{204}{42.857} ≈ 4.76 \ A$
Since resistance stays the same, the current decreases proportionally with the voltage.
Step 4: Find the new resistance
$R_{new}=0.85 \times 42.857 ≈ 4.76 \ A$
A 15% reduction means the resistance becomes 85% of its original value.
Step 5: Find the new current
$I=\frac{V}{R_{new}}$
$I = \frac{240}{36.43} ≈ 6.59 \ A$
With lower resistance, current increases because it is easier for charge to flow.
Applications of Resistance, Ohm’s Law, and Electrical Conductivity
The concepts of resistance, Ohm’s law, and the relationship between potential difference (p.d.) and current are widely used in modern technology and everyday applications. These principles also connect with the microscopic relationship J = σE, which explains how current flows at the material level.
1. Electrical Circuit Design
Ohm’s law (V = IR) is essential in designing electrical circuits. Engineers use it to determine the correct values of voltage, current, and resistance to ensure circuits function safely and efficiently.
2. Power Distribution Systems
In power transmission, minimizing resistance is important to reduce energy loss. Conductors with high conductivity (low resistance), such as copper and aluminum, are used to transmit electricity over long distances.
3. Electronic Devices
Devices such as smartphones, computers, and televisions rely on controlled current flow. Resistors are used to regulate current, while the relationship between p.d. and current ensures proper device operation.
4. Heating Appliances
Appliances like electric heaters, irons, and kettles use resistance to produce heat. When current flows through a high-resistance material, electrical energy is converted into heat energy.
5. Material Science and Conductivity
The equation J = σE explains how current density depends on the electric field and the conductivity of a material. This is important in developing new materials such as semiconductors and superconductors.
6. Sensors and Measurement Systems
Many sensors operate based on changes in resistance or conductivity. For example, temperature sensors and strain gauges detect variations in electrical properties to measure physical changes.
Conclusion
Resistance, Ohm’s law, and the relationship between potential difference and current are essential concepts in electricity. These principles describe how electric current flows in a conducting material.
Ohm’s law (V = IR) provides a simple macroscopic relationship between voltage, current, and resistance, while the equation J = σE explains the microscopic behavior of charge carriers within a material.
Understanding these relationships is crucial in physics and engineering, as they are widely applied in electrical systems, electronic devices, and modern technological developments.
Frequently Asked Questions (FAQ)
What is resistance in electricity?
Resistance is the opposition offered by a material to the flow of electric current. It is measured in ohms (Ω).
What is Ohm’s law?
Ohm’s law states that the current flowing through a conductor is directly proportional to the potential difference across it, provided temperature remains constant. It is expressed as V = IR.
What is the relationship between potential difference and current?
The relationship is linear for an ohmic conductor, meaning current increases proportionally with potential difference, as described by Ohm’s law.
What does the equation J = σE represent?
The equation J = σE shows that current density (J) is directly proportional to the electric field (E), where σ is the electrical conductivity of the material.
How is Ohm’s law related to J = σE?
Ohm’s law (V = IR) describes the macroscopic behavior of current in a circuit, while J = σE describes the microscopic behavior within a material. Both equations are equivalent and represent the same physical principle.
Keywords: resistance FAQ, Ohm's law questions, current and voltage relationship, J = sigma E explanation, electrical conductivity FAQ
Post a Comment for "Ohm’s Law and Resistance Explained: Formula, Examples, and Applications"