To find the equivalent resistance between two points in a circuit, you need to combine the resistances of all components between those points. The method for calculating the equivalent resistance depends on whether the resistors are connected in series, parallel, or a combination of both. Let's break down the steps for each case:
### 1. **Resistors in Series**
When resistors are connected end-to-end, so that the same current flows through each resistor, they are said to be in series. The total or **equivalent resistance** (\(R_{eq}\)) of resistors in series is the sum of their individual resistances.
#### Formula:
\[
R_{eq} = R_1 + R_2 + R_3 + \ldots + R_n
\]
#### Example:
If you have three resistors in series with resistances of 2 Ω, 3 Ω, and 5 Ω:
\[
R_{eq} = 2 + 3 + 5 = 10 \, \Omega
\]
### 2. **Resistors in Parallel**
When resistors are connected in such a way that they have the same voltage across them but can carry different currents, they are in parallel. The equivalent resistance of resistors in parallel is found by taking the reciprocal of the sum of the reciprocals of their individual resistances.
#### Formula:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}
\]
Or, equivalently:
\[
R_{eq} = \frac{1}{\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}\right)}
\]
#### Example:
If you have three resistors in parallel with resistances of 2 Ω, 3 Ω, and 6 Ω:
\[
\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}
\]
Calculating the above:
\[
\frac{1}{R_{eq}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1
\]
Therefore:
\[
R_{eq} = 1 \, \Omega
\]
### 3. **Combination of Series and Parallel Resistors**
In more complex circuits, some resistors might be in series and others in parallel. In these cases, you need to **reduce** the circuit step-by-step by identifying and simplifying sections of series and parallel resistors.
#### Step-by-Step Approach:
1. **Identify Simplifiable Groups:** Look for groups of resistors that are clearly in series or parallel. Simplify them using the appropriate formula.
2. **Redraw the Circuit:** After simplifying a group, redraw the circuit to show the reduced equivalent resistance for that group.
3. **Repeat Until Fully Simplified:** Continue identifying groups of series or parallel resistors, simplify them, and redraw the circuit until you have a single equivalent resistance between the two points.
#### Example:
Consider a circuit with four resistors:
- \(R_1 = 4 \, \Omega\) and \(R_2 = 6 \, \Omega\) are in series.
- \(R_3 = 3 \, \Omega\) and \(R_4 = 6 \, \Omega\) are in parallel.
The combination of \(R_1\) and \(R_2\) in series is:
\[
R_{12} = R_1 + R_2 = 4 + 6 = 10 \, \Omega
\]
Next, the equivalent resistance of \(R_3\) and \(R_4\) in parallel is:
\[
\frac{1}{R_{34}} = \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}
\]
Thus:
\[
R_{34} = 2 \, \Omega
\]
Now, if \(R_{12}\) (10 Ω) and \(R_{34}\) (2 Ω) are in series, the total equivalent resistance is:
\[
R_{eq} = R_{12} + R_{34} = 10 + 2 = 12 \, \Omega
\]
### 4. **Wheatstone Bridge and More Complex Circuits**
For more complex circuits, such as those containing a **Wheatstone bridge** or where resistors form more complicated patterns, other methods such as:
- **Kirchhoff’s Rules** (Kirchhoff’s Current Law and Kirchhoff’s Voltage Law),
- **Node Voltage Method**, or
- **Mesh Current Method**,
may be needed to solve for equivalent resistance.
### Conclusion
To find the equivalent resistance between two points in a circuit:
1. Identify resistors in series and parallel.
2. Use the appropriate formulas to simplify series and parallel combinations.
3. Redraw and simplify the circuit step-by-step until only one equivalent resistance remains.
If you encounter more complex circuits, use Kirchhoff’s laws or more advanced techniques to solve them.