Parallel and combination circuit diagrams with resistor values, voltage sources, and current measurements for analysis.

Let's solve each of the parallel and combination circuits step by step using basic principles of Ohm’s Law and parallel/series resistance rules.

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🔧 Key Concepts:

- In parallel circuits, voltage across each branch is the same.
- Total current $ I_T = I_1 + I_2 + \dots $
- Equivalent resistance for parallel resistors:
$$
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots
$$
- Ohm’s Law: $ V = IR $

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## ✔ Circuit 1 (Top Left)

Given:
- Voltage source: 24 V
- $ R_1 = 12\,\Omega $, $ R_2 = 24\,\Omega $, in parallel

Step 1: Find $ R_{eq} $

$$
\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}
\Rightarrow R_{eq} = 8\,\Omega
$$

Step 2: Total Current $ I_T $

$$
I_T = \frac{V}{R_{eq}} = \frac{24}{8} = 3\,\text{A}
$$

Step 3: Voltages

In parallel, $ V_1 = V_2 = V_T = 24\,\text{V} $

Step 4: Individual Currents

$$
I_1 = \frac{V_1}{R_1} = \frac{24}{12} = 2\,\text{A} \\
I_2 = \frac{V_2}{R_2} = \frac{24}{24} = 1\,\text{A}
$$

✔ Answer:
- $ R_{eq} = 8\,\Omega $
- $ I_T = 3\,\text{A} $
- $ V_1 = 24\,\text{V} $
- $ V_2 = 24\,\text{V} $
- $ I_1 = 2\,\text{A} $
- $ I_2 = 1\,\text{A} $

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## ✔ Circuit 2 (Top Right)

Given:
- Total current $ I_T = 4\,\text{A} $
- $ R_1 = 20\,\Omega $, $ R_2 = 20\,\Omega $, in parallel

Step 1: $ R_{eq} $

$$
\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \Rightarrow R_{eq} = 10\,\Omega
$$

Step 2: $ V_T $ (Total Voltage)

$$
V_T = I_T \cdot R_{eq} = 4 \times 10 = 40\,\text{V}
$$

Step 3: Voltages

$ V_1 = V_2 = V_T = 40\,\text{V} $

Step 4: Currents

Since both resistors are equal:
$$
I_1 = I_2 = \frac{V}{R} = \frac{40}{20} = 2\,\text{A}
$$

✔ Answer:
- $ R_{eq} = 10\,\Omega $
- $ I_T = 4\,\text{A} $
- $ V_T = 40\,\text{V} $
- $ V_1 = 40\,\text{V} $
- $ I_1 = 2\,\text{A} $
- $ I_2 = 2\,\text{A} $

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## ✔ Circuit 3 (Bottom Left)

Given:
- Voltage source: 6 V
- Two branches with ammeters showing $ I_1 = 2\,\text{A} $, $ I_2 = 1\,\text{A} $

Step 1: Voltages

In parallel: $ V_1 = V_2 = 6\,\text{V} $

Step 2: Resistances

Use Ohm’s Law:
$$
R_1 = \frac{V_1}{I_1} = \frac{6}{2} = 3\,\Omega \\
R_2 = \frac{V_2}{I_2} = \frac{6}{1} = 6\,\Omega
$$

Step 3: $ R_{eq} $

$$
\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \Rightarrow R_{eq} = 2\,\Omega
$$

Step 4: $ I_T $

$$
I_T = I_1 + I_2 = 2 + 1 = 3\,\text{A}
$$

✔ Answer:
- $ V_1 = 6\,\text{V} $
- $ V_2 = 6\,\text{V} $
- $ R_1 = 3\,\Omega $
- $ R_2 = 6\,\Omega $
- $ R_{eq} = 2\,\Omega $
- $ I_1 = 2\,\text{A} $
- $ I_2 = 1\,\text{A} $
- $ I_T = 3\,\text{A} $

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## ✔ Circuit 4 (Bottom Right)

Given:
- Voltage source: 40 V
- $ R_1 = 5\,\Omega $, $ R_2 = 20\,\Omega $, $ R_3 = 4\,\Omega $ — all in parallel

Step 1: $ R_{eq} $

$$
\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{20} + \frac{1}{4}
= \frac{4}{20} + \frac{1}{20} + \frac{5}{20} = \frac{10}{20} = \frac{1}{2}
\Rightarrow R_{eq} = 2\,\Omega
$$

Step 2: $ I_T $

$$
I_T = \frac{V}{R_{eq}} = \frac{40}{2} = 20\,\text{A}
$$

Step 3: Voltages

All branches have same voltage:
$ V_1 = V_2 = V_3 = 40\,\text{V} $

Step 4: Individual Currents

$$
I_1 = \frac{40}{5} = 8\,\text{A} \\
I_2 = \frac{40}{20} = 2\,\text{A} \\
I_3 = \frac{40}{4} = 10\,\text{A}
$$

Check: $ 8 + 2 + 10 = 20\,\text{A} $ → Correct

✔ Answer:
- $ R_{eq} = 2\,\Omega $
- $ I_T = 20\,\text{A} $
- $ V_1 = 40\,\text{V} $
- $ V_2 = 40\,\text{V} $
- $ V_3 = 40\,\text{V} $
- $ I_1 = 8\,\text{A} $
- $ I_2 = 2\,\text{A} $
- $ I_3 = 10\,\text{A} $

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## ✔ Combination Circuit (Bottom)

Circuit Description:
- 12 V battery
- A 2 Ω resistor in series with a parallel combination of two 2 Ω resistors

Label:
- $ R_1 = 2\,\Omega $: series resistor
- $ R_2 = 2\,\Omega $, $ R_3 = 2\,\Omega $: in parallel

We need to fill this table:

| | V | I | R |
|-------|-------|-------|-------|
| R1 | | | 2 Ω |
| R2 | | | 2 Ω |
| R3 | | | 2 Ω |
| Total | 12 V | | |

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Step 1: Find equivalent resistance

#### Parallel part ($ R_2 $ and $ R_3 $):
$$
\frac{1}{R_{23}} = \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow R_{23} = 1\,\Omega
$$

#### Total resistance:
Series with $ R_1 $:
$$
R_{total} = R_1 + R_{23} = 2 + 1 = 3\,\Omega
$$

Step 2: Total current $ I_T $

$$
I_T = \frac{V_{total}}{R_{total}} = \frac{12}{3} = 4\,\text{A}
$$

This current flows through $ R_1 $ only.

Step 3: Voltage drop across $ R_1 $

$$
V_1 = I_T \cdot R_1 = 4 \times 2 = 8\,\text{V}
$$

Step 4: Voltage across parallel combo

Remaining voltage:
$$
V_{23} = 12 - 8 = 4\,\text{V}
\Rightarrow V_2 = V_3 = 4\,\text{V}
$$

Step 5: Currents through $ R_2 $ and $ R_3 $

Since both are 2 Ω and have 4 V:
$$
I_2 = \frac{4}{2} = 2\,\text{A} \\
I_3 = \frac{4}{2} = 2\,\text{A}
$$

Check: $ I_2 + I_3 = 4\,\text{A} $ → matches $ I_T $

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Final Table:

| | V | I | R |
|-------|-------|-------|-------|
| R1 | 8 V | 4 A | 2 Ω |
| R2 | 4 V | 2 A | 2 Ω |
| R3 | 4 V | 2 A | 2 Ω |
| Total | 12 V | 4 A | 3 Ω |

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## ✔ Summary of All Answers

Circuit 1:

- $ R_{eq} = 8\,\Omega $
- $ I_T = 3\,\text{A} $
- $ V_1 = 24\,\text{V} $
- $ V_2 = 24\,\text{V} $
- $ I_1 = 2\,\text{A} $
- $ I_2 = 1\,\text{A} $

Circuit 2:

- $ R_{eq} = 10\,\Omega $
- $ I_T = 4\,\text{A} $
- $ V_T = 40\,\text{V} $
- $ V_1 = 40\,\text{V} $
- $ I_1 = 2\,\text{A} $
- $ I_2 = 2\,\text{A} $

Circuit 3:

- $ V_1 = 6\,\text{V} $
- $ V_2 = 6\,\text{V} $
- $ R_1 = 3\,\Omega $
- $ R_2 = 6\,\Omega $
- $ R_{eq} = 2\,\Omega $
- $ I_1 = 2\,\text{A} $
- $ I_2 = 1\,\text{A} $
- $ I_T = 3\,\text{A} $

Circuit 4:

- $ R_{eq} = 2\,\Omega $
- $ I_T = 20\,\text{A} $
- $ V_1 = 40\,\text{V} $
- $ V_2 = 40\,\text{V} $
- $ V_3 = 40\,\text{V} $
- $ I_1 = 8\,\text{A} $
- $ I_2 = 2\,\text{A} $
- $ I_3 = 10\,\text{A} $

Combination Circuit Table:

| | V | I | R |
|-------|-------|-------|-------|
| R1 | 8 V | 4 A | 2 Ω |
| R2 | 4 V | 2 A | 2 Ω |
| R3 | 4 V | 2 A | 2 Ω |
| Total | 12 V | 4 A | 3 Ω |

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Let me know if you'd like a diagram or explanation of any step!