Electric Current and Potential Difference
Electric circuits
Electric circuits are systems where electric charge flows through a closed loop.
To understand how these circuits work, we must first define two fundamental concepts: electric current and electric potential difference.
Electric Current: The Flow of Charge
Electric current
Electric current is the rate at which electric charge flows through a conductor.
- Consider water flowing through a pipe.
- The amount of water passing a point per second is similar to the amount of electric charge passing through a wire per second.
Mathematically, current ($I$) is expressed as:
$$I = \frac{\Delta q}{\Delta t}$$
where:
- $I$ is the current in amperes (A).
- $\Delta q$ is the change in charge in coulombs (C).
- $\Delta t$ is the change in time in seconds (s).
The ampere (A) is the unit of electric current, defined as one coulomb of charge passing through a point in a circuit per second.
Direct Current (DC)
- In direct current (DC), charge carriers (usually electrons) flow in a single direction.
- This is the type of current produced by batteries and used in most electronic devices.
If 6 coulombs of charge pass through a wire in 2 seconds, the current is: $$I = \frac{\Delta q}{\Delta t} = \frac{6 \, \mathrm{C}}{2 \, \mathrm{s}} = 3 \, \mathrm{A}$$
Calculating Current
To calculate current, you need to know the total charge passing through a point and the time it takes.
To find the number of electrons in a given charge, divide the total charge by the charge of a single electron ($1.6 \times 10^{-19}$ C).
A current of 1 A is established in a conductor. How many electrons move through the cross-sectional area of the conductor in 1 s?
Solution
- The charge of one electron is $1.6 \times 10^{-19} \, \mathrm{C}$.
- In 1 second, 1 coulomb of charge passes through the conductor.
- The number of electrons is: $$\frac{1 , \mathrm{C}}{1.6 \times 10^{-19} \, \mathrm{C/electron}} = 6.25 \times 10^{18} \, \mathrm{electrons}$$
Electric Potential Difference (Voltage)
Electric potential difference
Electric potential difference, or voltage, is the energy per unit charge required to move a charge between two points in an electric field.
Mathematically, voltage ($V$) is expressed as:
$$V = \frac{W}{q}$$
where:
- $V$ is the potential difference in volts (V).
- $W$ is the work done in joules (J).
- $q$ is the charge in coulombs (C).
The volt (V) is the unit of potential difference, defined as one joule of work done per coulomb of charge.
Understanding Voltage
- Voltage can be thought of as the "push" that drives charge through a circuit.
- It is created by a source of energy, such as a battery, which does work to move charges from a lower potential to a higher potential.
The work done in moving a charge of $2.0 \, \mu\mathrm{C}$ between two points in an electric field is $1.50 \times 10^{-4} \, \mathrm{J}$.
Determine the potential difference between the two points.
Solution
- Use the formula $V = \frac{W}{q}$.
- Substitute the values:$$V = \frac{1.50 \times 10^{-4} \, \mathrm{J}}{2.0 \times 10^{-6} \, \mathrm{C}} = 75 \, \mathrm{V}$$
Voltage and Energy
- Voltage is directly related to the energy transferred to or from a charge.
- When a charge moves through a potential difference, it gains or loses energy.
An electron is accelerated from rest through a potential difference of 75 V. What is the speed acquired by the electron?
Solution
- The work done on the electron is $W = qV$, where $q = 1.6 \times 10^{-19} \, \mathrm{C}$ (the charge of an electron).
- The work done is converted into kinetic energy: $$W = \frac{1}{2}mv^2$$
- Substitute the values: $$1.6 \times 10^{-19} \times 75 = \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2$$
- Solve for $v$: $$v = \sqrt{\frac{2 \times 1.2 \times 10^{-17}}{9.1 \times 10^{-31}}} = 5.1 \times 10^6 \, \mathrm{ms}^{-1}$$
Connecting Current and Voltage
In a circuit, current and voltage are closely related.
Voltage provides the energy needed to move charges, while current is the flow of those charges.
Calculating Power in a Circuit
Power in a circuit
The power dissipated in a circuit component is the rate at which energy is transferred.
It can be calculated using the formula:
$$P = IV$$
where:
- $P$ is the power in watts (W).
- $I$ is the current in amperes (A).
- $V$ is the voltage in volts (V).
A resistor has a current of 2.0 A flowing through it and a potential difference of 12 V across its ends. What is the power dissipated in the resistor?
Solution
- Use the formula $P = IV$.
- Substitute the values:$$P = 2.0 \, \mathrm{A} \times 12 \, \mathrm{V} = 24 \, \mathrm{W}$$
- How is electric current defined, and what is its unit?
- What is the relationship between work, charge, and potential difference?
- How would you calculate the power dissipated in a resistor with a current of 3 A and a voltage of 9 V?
