George Meshveliani
Electric current ($I$) is the rate of flow of electric charge. Think of it as the volume of water flowing through a pipe per second.
Formula: $I = \frac{\Delta Q}{\Delta t}$
Unit: Ampere (A), where $1 \, \text{A} = 1 \, \text{C}/\text{s}$.
Voltage ($V$), or potential difference, is the energy per unit charge. It's the "push" that makes current flow, like the pressure difference in a water pipe.
Formula: $V = \frac{\Delta E}{\Delta Q}$
Unit: Volt (V), where $1 \, \text{V} = 1 \, \text{J}/\text{C}$.
Resistance ($R$) is the opposition to the flow of electric current. In a water pipe, a narrower section offers more resistance to the water flow.
Formula: $R = \rho \frac{L}{A}$
Unit: Ohm ($\Omega$).
Ohm's Law describes the relationship between voltage, current, and resistance in a circuit.
$$V = IR$$
This can be rearranged to solve for any variable:
Components are connected one after another, forming a single path for current to flow.
Components are connected across the same two points, creating multiple paths for current.
The slope of a V-I graph represents the resistance of the component. A steeper slope means higher resistance.
A series circuit has a $12 \, \text{V}$ battery and two resistors with values of $4 \, \Omega$ and $2 \, \Omega$. Calculate the total resistance of the circuit and the current flowing through it.
Hint: Remember to add resistances in series, then use Ohm's Law.
First, find the total resistance ($R_{total}$):
$$R_{total} = R_1 + R_2 = 4 + 2 = 6 \, \Omega$$
Now, use Ohm's Law to find the total current ($I_{total}$):
$$I_{total} = \frac{V_{total}}{R_{total}} = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A}$$
A parallel circuit has a $12 \, \text{V}$ battery and two resistors with values of $3 \, \Omega$ and $6 \, \Omega$. Calculate the total resistance of the circuit and the total current flowing through the circuit.
Hint: Use the reciprocal formula for parallel resistance, then use Ohm's Law.
First, find the total resistance ($R_{total}$):
$$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$
$$R_{total} = 2 \, \Omega$$
Now, use Ohm's Law to find the total current ($I_{total}$):
$$I_{total} = \frac{V_{total}}{R_{total}} = \frac{12 \, \text{V}}{2 \, \Omega} = 6 \, \text{A}$$
A circuit has a $20 \, \text{V}$ battery. The circuit contains a $5 \, \Omega$ resistor in series with a parallel combination of two $10 \, \Omega$ resistors. Find the total resistance and the total current flowing from the battery.
Hint: Simplify the parallel part of the circuit first, then add the series resistance.
First, find the equivalent resistance of the two parallel resistors ($R_p$):
$$\frac{1}{R_{p}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$$
$$R_p = 5 \, \Omega$$
Now, add this parallel resistance to the series resistor to find the total resistance ($R_{total}$):
$$R_{total} = R_{series} + R_p = 5 + 5 = 10 \, \Omega$$
Finally, use Ohm's Law to find the total current ($I_{total}$):
$$I_{total} = \frac{V_{total}}{R_{total}} = \frac{20 \, \text{V}}{10 \, \Omega} = 2 \, \text{A}$$
Any questions?