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Series and Parallel Circuits

CSCA Series and Parallel Circuits study guide organized around the publicly available CSCA syllabus. Practice Physics questions on aicsca.com.

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Topic: Electromagnetism > DC Circuits > Series and Parallel Circuits

1. Core Model Comparison: Series vs Parallel

Distinguishing the connection method is the starting point for all circuit problems. Use the diagram below to intuitively understand the difference in current paths.

| Feature | **Series Circuit** | **Parallel Circuit** |

| :--- | :--- | :--- |

| **Connection** | **End-to-End**, no branches | **Head-to-Head, Tail-to-Tail**, branched |

| **Current Path** | Only **ONE** path | **MULTIPLE** paths (Main + Branches) |

| **Break Effect** | One break, **ALL stop** | One branch break, **Others UNAFFECTED** |

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2. Core Quantitative Rules

This is the foundation for calculation problems in the CSCA exam.

(1) Series Circuit

* **Current**: Same everywhere. $$I = I_1 = I_2$$

* **Voltage**: Total voltage equals sum of individual voltages. $$U = U_1 + U_2$$

* **Resistance**: Total resistance equals sum of individual resistances (Increases). $$R_{eq} = R_1 + R_2$$

* **Key Inference: Voltage Divider**

Voltage is **proportional** to resistance:

$$\frac{U_1}{U_2} = \frac{R_1}{R_2} \quad \Rightarrow \quad U_1 = \frac{R_1}{R_1 + R_2} U$$

(2) Parallel Circuit

* **Voltage**: Same across all branches. $$U = U_1 = U_2$$

* **Current**: Main current equals sum of branch currents. $$I = I_1 + I_2$$

* **Resistance**: Reciprocal of total R equals sum of reciprocals (Decreases).

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$

**Exam Tip**: For **two** resistors in parallel, use the shortcut: $$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$$

* **Key Inference: Current Divider**

Current is **inversely proportional** to resistance:

$$\frac{I_1}{I_2} = \frac{R_2}{R_1} \quad \Rightarrow \quad I_1 = \frac{R_2}{R_1 + R_2} I$$

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3. Typical Examples

**Example 1: Switch Changing Circuit Structure**

**Problem**: As shown, $R_1 = 2\Omega$, $R_2 = 3\Omega$, $R_3 = 6\Omega$. Find the total resistance when:

(1) Switch S is OPEN.

(2) Switch S is CLOSED.

**Solution**:

1. **S Open**: $R_3$ branch is open (no current). Circuit is $R_1$ and $R_2$ in **Series**.

$$R_{total} = R_1 + R_2 = 2 + 3 = 5\Omega$$

2. **S Closed**: Current flows through $R_1$, then splits into $R_2$ and $R_3$. So, $R_2$ and $R_3$ are in **Parallel**, and this block is in **Series** with $R_1$.

* Parallel part: $R_{23} = \frac{3 \times 6}{3 + 6} = 2\Omega$

* Total: $R_{total} = R_1 + R_{23} = 2 + 2 = 4\Omega$

**Example 2: Mixed Circuit & Ammeter**

**Problem**: Source $U=12V$, $R_1=4\Omega$, $R_2=6\Omega$, $R_3=12\Omega$. Switch S controls the branch with $R_2$. Find the Ammeter reading (main current) when:

(1) S is OPEN.

(2) S is CLOSED.

**Solution**:

1. **S Open**: $R_2$ is disconnected. Circuit is $R_1$ and $R_3$ in **Series**.

$$I = \frac{U}{R_1 + R_3} = \frac{12}{4 + 12} = 0.75A$$

2. **S Closed**: $R_2$ and $R_3$ in **Parallel**, then in **Series** with $R_1$.

* Parallel R: $R_{23} = \frac{6 \times 12}{6 + 12} = 4\Omega$

* Total R: $R_{total} = R_1 + R_{23} = 4 + 4 = 8\Omega$

* Current: $I' = \frac{U}{R_{total}} = \frac{12}{8} = 1.5A$

**Example 3: Current Divider Application**

**Problem**: $R_1=10\Omega$ in parallel with $R_2=30\Omega$. Total current $I=0.8A$. Find current $I_1$ through $R_1$.

**Solution**:

Using **Current Divider Rule** (Current splits inversely to resistance):

$$I_1 = \frac{R_2}{R_1 + R_2} \times I$$

$$I_1 = \frac{30}{10 + 30} \times 0.8 = 0.6A$$

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4. CSCA Common Pitfalls

1. **Parallel R Calculation**: Don't forget to invert the result ($1/R$). Check: **Total Parallel R < Smallest Branch R**.

2. **Circuit ID**: For complex circuits, trace the **current path**. Treat ideal Voltmeters as open gaps and Ammeters as wires.

3. **Formula Confusion**: Voltage divides proportionally ($R \uparrow \Rightarrow U \uparrow$), Current divides inversely ($R \uparrow \Rightarrow I \downarrow$).