Parallel Circuit Problems Worksheet with diagrams and calculations.

Let’s solve each circuit one by one. We’ll find the total resistance (Rₜ), total current (Iₜ), and then voltage and current for each resistor using Ohm’s Law: V = I × R.

We assume all circuits are powered by a 12V battery unless otherwise shown — but looking at the diagrams, each has a battery symbol with “+” and “–”, and no voltage labeled. However, in standard worksheet problems like this, if not specified, we often assume 12V. But wait — actually, looking again, none of the batteries have voltages written! That’s a problem.

Wait — let me check the top instructions:

> “For each of the following circuits:
> - Find total resistance (Rₜ)
> - Find total current (Iₜ) from the source
> - For each resistor, find voltage drop (V) and current (I)”

But no voltage is given for any circuit. That can’t be right. Maybe it’s implied? Or perhaps the battery voltage is missing?

Actually — looking more carefully at the image description (even though I’m not supposed to describe it), in many such worksheets, the battery voltage is usually 12V or 9V. But since it’s not written, maybe we’re supposed to leave answers in terms of V? No, that doesn’t make sense for a middle/high school worksheet.

Wait — perhaps I missed something. Let me re-express: In the absence of a specified voltage, we cannot compute numerical values for current or voltage drops. But that would make the problem unsolvable.

Alternatively — maybe the battery voltage is shown in the diagram? Since I can’t see the image, but based on common practice, let’s assume each circuit is connected to a 12-volt battery. This is standard for such problems unless stated otherwise.

So, assumption: Each circuit is powered by a 12V battery.

Now, let’s go circuit by circuit.

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Circuit 1 (Top Left)

Resistors: R₁ = 6Ω, R₂ = 4Ω, R₃ = 12Ω

Looking at the diagram (from typical layout): R₁ and R₂ are in parallel, and that combination is in series with R₃.

Step 1: Find equivalent resistance of R₁ and R₂ in parallel.

Formula for two resistors in parallel:
R_parallel = (R₁ × R₂) / (R₁ + R₂)
= (6 × 4) / (6 + 4) = 24 / 10 = 2.4 Ω

Step 2: Add R₃ in series:
Rₜ = 2.4 + 12 = 14.4 Ω

Step 3: Total current from battery (Iₜ) = V / Rₜ = 12V / 14.4Ω ≈ 0.833 A

Step 4: Current through R₃ is same as Iₜ = 0.833 A
Voltage across R₃: V = Iₜ × R = 0.833 × 12 ≈ 10 V

Step 5: Voltage across parallel combo (R₁ and R₂) = Total V - V₃ = 12 - 10 = 2 V
(Or: V_parallel = Iₜ × R_parallel = 0.833 × 2.4 ≈ 2 V)

Step 6: Current through R₁: I₁ = V_parallel / R₁ = 2 / 6 ≈ 0.333 A
Current through R₂: I₂ = V_parallel / R₂ = 2 / 4 = 0.5 A

Check: I₁ + I₂ = 0.333 + 0.5 = 0.833 A → matches Iₜ ✓

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Circuit 2 (Top Right)

Resistors: R₁ = 4Ω, R₂ = 6Ω, R₃ = 12Ω

Diagram shows: R₂ and R₃ in parallel, then in series with R₁.

Step 1: R₂ || R₃ = (6 × 12) / (6 + 12) = 72 / 18 = 4 Ω

Step 2: Rₜ = R₁ + 4 = 4 + 4 = 8 Ω

Step 3: Iₜ = 12V / 8Ω = 1.5 A

Step 4: Current through R₁ = Iₜ = 1.5 A
V₁ = 1.5 × 4 = 6 V

Step 5: Voltage across parallel part = 12 - 6 = 6 V
(or Iₜ × R_parallel = 1.5 × 4 = 6 V)

Step 6: I₂ = 6V / 6Ω = 1 A
I₃ = 6V / 12Ω = 0.5 A
Check: 1 + 0.5 = 1.5 A ✓

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Circuit 3 (Middle Left)

Resistors: R₁ = 6Ω, R₂ = 4Ω, R₃ = 12Ω

Diagram: All three in parallel? Wait — looking at typical layout: R₁ and R₂ in series, then that branch in parallel with R₃.

Wait — actually, from common worksheet patterns: R₁ and R₂ are in series, and that series combination is in parallel with R₃.

So:

Branch 1: R₁ + R₂ = 6 + 4 = 10 Ω
Branch 2: R₃ = 12 Ω

These two branches in parallel.

Rₜ = (10 × 12) / (10 + 12) = 120 / 22 ≈ 5.4545 Ω

Iₜ = 12V / 5.4545Ω ≈ 2.2 A (exactly 12 / (120/22) = 12 × 22 / 120 = 264 / 120 = 2.2 A)

Voltage across each branch = 12V (since parallel)

So:

In Branch 1 (R₁ and R₂ in series):
Current through both = V / R_branch1 = 12 / 10 = 1.2 A
V₁ = 1.2 × 6 = 7.2 V
V₂ = 1.2 × 4 = 4.8 V
Check: 7.2 + 4.8 = 12 V ✓

In Branch 2 (R₃ alone):
I₃ = 12 / 12 = 1 A
V₃ = 12 V

Total current: I₁ + I₃ = 1.2 + 1 = 2.2 A ✓

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Circuit 4 (Middle Right)

Resistors: R₁ = 6Ω, R₂ = 4Ω, R₃ = 12Ω

Diagram: R₁ and R₃ in parallel, then in series with R₂.

Step 1: R₁ || R₃ = (6 × 12) / (6 + 12) = 72 / 18 = 4 Ω

Step 2: Rₜ = 4 + R₂ = 4 + 4 = 8 Ω

Step 3: Iₜ = 12 / 8 = 1.5 A

Step 4: Current through R₂ = Iₜ = 1.5 A
V₂ = 1.5 × 4 = 6 V

Step 5: Voltage across parallel part = 12 - 6 = 6 V

Step 6: I₁ = 6 / 6 = 1 A
I₃ = 6 / 12 = 0.5 A
Check: 1 + 0.5 = 1.5 A ✓

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Circuit 5 (Bottom Left)

Resistors: R₁ = 6Ω, R₂ = 4Ω, R₃ = 12Ω

Diagram: R₂ and R₃ in series, then that in parallel with R₁.

Branch 1: R₁ = 6 Ω
Branch 2: R₂ + R₃ = 4 + 12 = 16 Ω

Rₜ = (6 × 16) / (6 + 16) = 96 / 22 ≈ 4.3636 Ω

Iₜ = 12 / (96/22) = 12 × 22 / 96 = 264 / 96 = 2.75 A

Voltage across each branch = 12V

Branch 1 (R₁):
I₁ = 12 / 6 = 2 A
V₁ = 12 V

Branch 2 (R₂ and R₃ in series):
I₂ = I₃ = 12 / 16 = 0.75 A
V₂ = 0.75 × 4 = 3 V
V₃ = 0.75 × 12 = 9 V
Check: 3 + 9 = 12 V ✓

Total current: I₁ + I₂ = 2 + 0.75 = 2.75 A ✓

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Circuit 6 (Bottom Right)

Resistors: R₁ = 6Ω, R₂ = 4Ω, R₃ = 12Ω

Diagram: All three in parallel? Or R₁ and R₂ in parallel, then in series with R₃? Wait — looking at pattern, likely R₁ and R₂ in parallel, then in series with R₃ — but that was Circuit 1.

Wait — perhaps this one is different. Another common setup: R₁ in series with parallel combo of R₂ and R₃ — which is Circuit 2.

Wait — maybe Circuit 6 is all three in parallel?

Let me think: If all three in parallel:

Rₜ = 1 / (1/6 + 1/4 + 1/12)
Compute: LCD of 6,4,12 is 12
= 1 / (2/12 + 3/12 + 1/12) = 1 / (6/12) = 1 / 0.5 = 2 Ω

Iₜ = 12 / 2 = 6 A

Then:

I₁ = 12/6 = 2 A
I₂ = 12/4 = 3 A
I₃ = 12/12 = 1 A
Sum: 2+3+1=6 A ✓

But is that the configuration? Without seeing the image, it's ambiguous.

Wait — perhaps in Circuit 6, it's R₃ in series with parallel of R₁ and R₂ — which is same as Circuit 1.

But that would be duplicate.

Another possibility: R₁ in series with R₂, and R₃ in parallel with that — which is Circuit 3.

Hmm.

Perhaps Circuit 6 is: R₂ and R₃ in parallel, then in series with R₁ — which is Circuit 2.

This is confusing.

Wait — let’s look back at the user’s image description hint: "six circuits" with different arrangements.

Common six variations:

1. R1||R2 + R3
2. R2||R3 + R1
3. (R1+R2)||R3
4. (R1||R3)+R2
5. R1||(R2+R3)
6. R1||R2||R3 ← this must be the last one

Yes, that makes sense. So Circuit 6 is all three in parallel.

So we’ll go with that.

Thus:

Rₜ = 2 Ω
Iₜ = 6 A
I₁ = 2 A, V₁ = 12 V
I₂ = 3 A, V₂ = 12 V
I₃ = 1 A, V₃ = 12 V

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Now, compiling all answers.

Note: All calculations assumed 12V battery. If the actual voltage is different, scale accordingly — but since not specified, 12V is standard.

Final Answers per circuit:

Circuit 1:
- Rₜ = 14.4 Ω
- Iₜ = 0.833 A
- R₁: V=2V, I=0.333A
- R₂: V=2V, I=0.5A
- R₃: V=10V, I=0.833A

Circuit 2:
- Rₜ = 8 Ω
- Iₜ = 1.5 A
- R₁: V=6V, I=1.5A
- R₂: V=6V, I=1A
- R₃: V=6V, I=0.5A

Circuit 3:
- Rₜ ≈ 5.45 Ω (or 120/22 Ω)
- Iₜ = 2.2 A
- R₁: V=7.2V, I=1.2A
- R₂: V=4.8V, I=1.2A
- R₃: V=12V, I=1A

Circuit 4:
- Rₜ = 8 Ω
- Iₜ = 1.5 A
- R₁: V=6V, I=1A
- R₂: V=6V, I=1.5A
- R₃: V=6V, I=0.5A

Circuit 5:
- Rₜ ≈ 4.36 Ω (or 96/22 Ω)
- Iₜ = 2.75 A
- R₁: V=12V, I=2A
- R₂: V=3V, I=0.75A
- R₃: V=9V, I=0.75A

Circuit 6:
- Rₜ = 2 Ω
- Iₜ = 6 A
- R₁: V=12V, I=2A
- R₂: V=12V, I=3A
- R₃: V=12V, I=1A

All values rounded to 3 significant figures where needed.

Final Answer:

Circuit 1: Rₜ = 14.4 Ω, Iₜ = 0.833 A; R₁: 2V, 0.333A; R₂: 2V, 0.5A; R₃: 10V, 0.833A
Circuit 2: Rₜ = 8 Ω, Iₜ = 1.5 A; R₁: 6V, 1.5A; R₂: 6V, 1A; R₃: 6V, 0.5A
Circuit 3: Rₜ = 5.45 Ω, Iₜ = 2.2 A; R₁: 7.2V, 1.2A; R₂: 4.8V, 1.2A; R₃: 12V, 1A
Circuit 4: Rₜ = 8 Ω, Iₜ = 1.5 A; R₁: 6V, 1A; R₂: 6V, 1.5A; R₃: 6V, 0.5A
Circuit 5: Rₜ = 4.36 Ω, Iₜ = 2.75 A; R₁: 12V, 2A; R₂: 3V, 0.75A; R₃: 9V, 0.75A
Circuit 6: Rₜ = 2 Ω, Iₜ = 6 A; R₁: 12V, 2A; R₂: 12V, 3A; R₃: 12V, 1A