Parallel Circuit Problems Worksheet with diagrams and equations for educational use.

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 across branches is the same. For series, current is the same.

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Circuit 1 (Top Left)

Given:
- Battery = 24 V
- R₁ = 30 Ω, R₂ = 60 Ω, R₃ = 90 Ω — all in parallel
- Ammeter A₁ measures total current from battery
- Voltmeter V₁ measures voltage across R₁ (which equals battery voltage since it's parallel)

In parallel circuits:
→ Voltage across each resistor = battery voltage = 24 V

So:
- V₁ = 24 V

Now find currents through each resistor using I = V/R:

- I₁ = 24 V / 30 Ω = 0.8 A
- I₂ = 24 V / 60 Ω = 0.4 A
- I₃ = 24 V / 90 Ω ≈ 0.267 A

Total current (A₁) = I₁ + I₂ + I₃ = 0.8 + 0.4 + 0.267 = 1.467 A

But let’s keep fractions to be exact:

I₁ = 24/30 = 4/5 = 0.8 A
I₂ = 24/60 = 2/5 = 0.4 A
I₃ = 24/90 = 4/15 ≈ 0.2667 A

Total I = 4/5 + 2/5 + 4/15 = (12/15 + 6/15 + 4/15) = 22/15 ≈ 1.467 A

✔ Final values for Circuit 1:
- V₁ = 24 V
- I₁ = 0.8 A
- I₂ = 0.4 A
- I₃ = 0.267 A (or 4/15 A)
- A₁ = 1.467 A (or 22/15 A)

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Circuit 2 (Top Right)

Given:
- Battery = 12 V
- R₁ = 12 Ω, R₂ = 24 Ω, R₃ = 36 Ω — all in parallel
- A₁ = total current
- V₁ = voltage across R₁ = battery voltage = 12 V

Again, parallel → voltage same everywhere = 12 V

Currents:

- I₁ = 12 V / 12 Ω = 1 A
- I₂ = 12 V / 24 Ω = 0.5 A
- I₃ = 12 V / 36 Ω = 1/3 A ≈ 0.333 A

Total current A₁ = 1 + 0.5 + 0.333 = 1.833 A

Exact fraction: 1 + 1/2 + 1/3 = 6/6 + 3/6 + 2/6 = 11/6 ≈ 1.833 A

✔ Final values for Circuit 2:
- V₁ = 12 V
- I₁ = 1 A
- I₂ = 0.5 A
- I₃ = 0.333 A (or 1/3 A)
- A₁ = 1.833 A (or 11/6 A)

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Circuit 3 (Middle Left)

This is a mix! Let’s look carefully.

Battery = 12 V
R₁ = 10 Ω and R₂ = 20 Ω are in series with each other.
That combination is in parallel with R₃ = 30 Ω.

Voltmeter V₁ is across R₁ → so we need voltage drop across R₁.
Ammeter A₁ is measuring total current from battery.

Step 1: Find equivalent resistance of R₁ and R₂ in series:

R_series = R₁ + R₂ = 10 + 20 = 30 Ω

Now this 30 Ω is in parallel with R₃ = 30 Ω.

So total resistance R_total = (30 × 30) / (30 + 30) = 900 / 60 = 15 Ω

Total current from battery (A₁) = V / R_total = 12 V / 15 Ω = 0.8 A

Now, since the two branches have equal resistance (both 30 Ω), current splits equally:

→ Each branch gets half of total current: 0.8 A / 2 = 0.4 A

So current through R₁-R₂ branch = 0.4 A → that’s also current through R₁ and R₂ (since they’re in series).

Therefore:
- I₁ = 0.4 A
- I₂ = 0.4 A
- I₃ = 0.4 A (other branch)

Voltage across R₁ (V₁) = I₁ × R₁ = 0.4 A × 10 Ω = 4 V

Check: Voltage across R₂ = 0.4 × 20 = 8 V → total across series pair = 4 + 8 = 12 V → matches battery → good.

✔ Final values for Circuit 3:
- V₁ = 4 V
- I₁ = 0.4 A
- I₂ = 0.4 A
- I₃ = 0.4 A
- A₁ = 0.8 A

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Circuit 4 (Middle Right)

Battery = 24 V
R₁ = 12 Ω, R₂ = 24 Ω — in series
This series combo is in parallel with R₃ = 36 Ω

V₁ is across R₁ → need voltage drop on R₁
A₁ is total current

Step 1: Series part: R₁ + R₂ = 12 + 24 = 36 Ω

Parallel with R₃ = 36 Ω → again, both branches 36 Ω → total R = (36×36)/(36+36) = 1296/72 = 18 Ω

Total current A₁ = 24 V / 18 Ω = 1.333... A = 4/3 A

Since both branches have same resistance (36 Ω), current splits evenly:

Each branch gets (4/3)/2 = 2/3 A ≈ 0.667 A

So current through R₁-R₂ branch = 2/3 A → same through R₁ and R₂.

Thus:
- I₁ = 2/3 A ≈ 0.667 A
- I₂ = 2/3 A ≈ 0.667 A
- I₃ = 2/3 A ≈ 0.667 A

Voltage across R₁ (V₁) = I₁ × R₁ = (2/3) × 12 = 8 V

Check: Voltage across R₂ = (2/3) × 24 = 16 V → total across series = 8 + 16 = 24 V → correct.

✔ Final values for Circuit 4:
- V₁ = 8 V
- I₁ = 0.667 A (or 2/3 A)
- I₂ = 0.667 A (or 2/3 A)
- I₃ = 0.667 A (or 2/3 A)
- A₁ = 1.333 A (or 4/3 A)

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Circuit 5 (Bottom Left)

Battery = 12 V
R₁ = 6 Ω, R₂ = 12 Ω — in series
This series combo is in parallel with R₃ = 18 Ω

V₁ is across R₁ → need voltage drop on R₁
A₁ is total current

Series resistance: R₁ + R₂ = 6 + 12 = 18 Ω

Parallel with R₃ = 18 Ω → again, equal resistances → total R = (18×18)/(18+18) = 324/36 = 9 Ω

Total current A₁ = 12 V / 9 Ω = 1.333... A = 4/3 A

Current splits equally between two 18 Ω branches → each gets (4/3)/2 = 2/3 A ≈ 0.667 A

So current through R₁-R₂ branch = 2/3 A → same through R₁ and R₂.

Thus:
- I₁ = 2/3 A ≈ 0.667 A
- I₂ = 2/3 A ≈ 0.667 A
- I₃ = 2/3 A ≈ 0.667 A

Voltage across R₁ (V₁) = I₁ × R₁ = (2/3) × 6 = 4 V

Check: Voltage across R₂ = (2/3) × 12 = 8 V → total = 4 + 8 = 12 V → correct.

✔ Final values for Circuit 5:
- V₁ = 4 V
- I₁ = 0.667 A (or 2/3 A)
- I₂ = 0.667 A (or 2/3 A)
- I₃ = 0.667 A (or 2/3 A)
- A₁ = 1.333 A (or 4/3 A)

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Circuit 6 (Bottom Right)

Battery = 12 V
R₁ = 4 Ω, R₂ = 8 Ω — in series
This series combo is in parallel with R₃ = 12 Ω

V₁ is across R₁ → need voltage drop on R₁
A₁ is total current

Series resistance: R₁ + R₂ = 4 + 8 = 12 Ω

Parallel with R₃ = 12 Ω → again, equal → total R = (12×12)/(12+12) = 144/24 = 6 Ω

Total current A₁ = 12 V / 6 Ω = 2 A

Current splits equally → each branch gets 2 A / 2 = 1 A

So current through R₁-R₂ branch = 1 A → same through R₁ and R₂.

Thus:
- I₁ = 1 A
- I₂ = 1 A
- I₃ = 1 A

Voltage across R₁ (V₁) = I₁ × R₁ = 1 A × 4 Ω = 4 V

Check: Voltage across R₂ = 1 × 8 = 8 V → total = 4 + 8 = 12 V → correct.

✔ Final values for Circuit 6:
- V₁ = 4 V
- I₁ = 1 A
- I₂ = 1 A
- I₃ = 1 A
- A₁ = 2 A

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## ✔ FINAL ANSWER SUMMARY:

We'll list answers per circuit as requested: V₁, I₁, I₂, I₃, A₁

Circuit 1:
V₁ = 24 V
I₁ = 0.8 A
I₂ = 0.4 A
I₃ = 0.267 A
A₁ = 1.467 A

Circuit 2:
V₁ = 12 V
I₁ = 1 A
I₂ = 0.5 A
I₃ = 0.333 A
A₁ = 1.833 A

Circuit 3:
V₁ = 4 V
I₁ = 0.4 A
I₂ = 0.4 A
I₃ = 0.4 A
A₁ = 0.8 A

Circuit 4:
V₁ = 8 V
I₁ = 0.667 A
I₂ = 0.667 A
I₃ = 0.667 A
A₁ = 1.333 A

Circuit 5:
V₁ = 4 V
I₁ = 0.667 A
I₂ = 0.667 A
I₃ = 0.667 A
A₁ = 1.333 A

Circuit 6:
V₁ = 4 V
I₁ = 1 A
I₂ = 1 A
I₃ = 1 A
A₁ = 2 A

(Note: Decimal values rounded to 3 decimal places where needed; fractional equivalents provided in reasoning.)