Electrical circuit diagrams illustrating different resistor arrangements and voltage sources.

Let’s solve each circuit one by one. We’ll use Ohm’s Law:
Voltage (V) = Current (I) × Resistance (R) → V = I × R
And for parallel circuits, voltage is the same across all branches. For series circuits, current is the same through all components.

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Circuit 1:

- Battery: 9 V
- Two resistors in parallel: R₁ = 3 Ω, R₂ = 6 Ω

In parallel, voltage across each resistor = battery voltage = 9 V.

→ Current through R₁: I₁ = V / R₁ = 9 / 3 = 3 A
→ Current through R₂: I₂ = V / R₂ = 9 / 6 = 1.5 A

Total current from battery: I_total = I₁ + I₂ = 3 + 1.5 = 4.5 A

✔ Answers:
I₁ = 3 A, I₂ = 1.5 A, I_total = 4.5 A

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Circuit 2:

- Battery: 12 V
- Two resistors in series: R₁ = 4 Ω, R₂ = 8 Ω

In series, total resistance R_total = R₁ + R₂ = 4 + 8 = 12 Ω

Current from battery (same through both): I = V / R_total = 12 / 12 = 1 A

Voltage drop across R₁: V₁ = I × R₁ = 1 × 4 = 4 V
Voltage drop across R₂: V₂ = I × R₂ = 1 × 8 = 8 V

Check: 4 V + 8 V = 12 V ✔

✔ Answers:
I = 1 A, V₁ = 4 V, V₂ = 8 V

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Circuit 3:

- Battery: 10 V
- Three resistors: R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 5 Ω — all in series

Total resistance: R_total = 2 + 3 + 5 = 10 Ω

Current from battery (same everywhere): I = V / R_total = 10 / 10 = 1 A

Voltage drops:
V₁ = I × R₁ = 1 × 2 = 2 V
V₂ = I × R₂ = 1 × 3 = 3 V
V₃ = I × R₃ = 1 × 5 = 5 V

Check: 2 + 3 + 5 = 10 V ✔

✔ Answers:
I = 1 A, V₁ = 2 V, V₂ = 3 V, V₃ = 5 V

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Circuit 4:

- Battery: 6 V
- Two resistors in parallel: R₁ = 3 Ω, R₂ = 6 Ω

Same as Circuit 1 but with 6 V.

Voltage across each = 6 V

→ I₁ = 6 / 3 = 2 A
→ I₂ = 6 / 6 = 1 A
→ I_total = 2 + 1 = 3 A

✔ Answers:
I₁ = 2 A, I₂ = 1 A, I_total = 3 A

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Circuit 5:

- Battery: 12 V
- R₁ = 4 Ω and R₂ = 4 Ω in parallel, then that combination in series with R₃ = 4 Ω

First, find equivalent resistance of parallel part:

R_parallel = (R₁ × R₂) / (R₁ + R₂) = (4 × 4) / (4 + 4) = 16 / 8 = 2 Ω

Now total resistance: R_total = R_parallel + R₃ = 2 + 4 = 6 Ω

Total current from battery: I_total = V / R_total = 12 / 6 = 2 A

This current goes through R₃ → so I₃ = 2 A

Voltage drop across R₃: V₃ = I₃ × R = 2 × 4 = 8 V

Remaining voltage for parallel branch: 12 - 8 = 4 V

So voltage across R₁ and R₂ = 4 V each (since parallel)

→ I₁ = 4 / 4 = 1 A
→ I₂ = 4 / 4 = 1 A

Check: I₁ + I₂ = 1 + 1 = 2 A = I_total ✔

✔ Answers:
I_total = 2 A, I₁ = 1 A, I₂ = 1 A, I₃ = 2 A, V₃ = 8 V

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Circuit 6:

- Battery: 12 V
- R₁ = 6 Ω and R₂ = 3 Ω in parallel, then that combination in series with R₃ = 2 Ω

Parallel part: R_parallel = (6 × 3) / (6 + 3) = 18 / 9 = 2 Ω

Total resistance: R_total = 2 + 2 = 4 Ω

Total current: I_total = 12 / 4 = 3 A

This current goes through R₃ → I = 3 A

Voltage drop across R₃: V₃ = 3 × 2 = 6 V

Voltage left for parallel branch: 12 - 6 = 6 V

So voltage across R₁ and R₂ = 6 V each

→ I₁ = 6 / 6 = 1 A
→ I₂ = 6 / 3 = 2 A

Check: I₁ + I₂ = 1 + 2 = 3 A = I_total ✔

✔ Answers:
I_total = 3 A, I₁ = 1 A, I₂ = 2 A, I₃ = 3 A, V₃ = 6 V

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Final Answer:

Circuit 1: I₁ = 3 A, I₂ = 1.5 A, I_total = 4.5 A
Circuit 2: I = 1 A, V₁ = 4 V, V₂ = 8 V
Circuit 3: I = 1 A, V₁ = 2 V, V₂ = 3 V, V₃ = 5 V
Circuit 4: I₁ = 2 A, I₂ = 1 A, I_total = 3 A
Circuit 5: I_total = 2 A, I₁ = 1 A, I₂ = 1 A, I₃ = 2 A, V₃ = 8 V
Circuit 6: I_total = 3 A, I₁ = 1 A, I₂ = 2 A, I₃ = 3 A, V₃ = 6 V