V, R, and I in Parallel Circuit Worksheet for 10th - Higher Ed ... - Free Printable

Let’s solve this step by step.

We are given a parallel circuit with three branches:

- Branch 1: R₁ = 470 Ω
- Branch 2: R₂ = 330 Ω
- Branch 3: R₃ = 560 Ω

The voltage across each branch is the same in a parallel circuit, and we’re told that V₁ = 9.8 V (voltage across R₁). Since it’s parallel, V₂ = V₃ = V_total = 9.8 V.

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Step 1: Find total resistance (R_T)

In a parallel circuit, total resistance is found using:

> 1/R_T = 1/R₁ + 1/R₂ + 1/R₃

Plug in the values:

1/R_T = 1/470 + 1/330 + 1/560

Let’s calculate each term:

- 1/470 ≈ 0.0021277
- 1/330 ≈ 0.0030303
- 1/560 ≈ 0.0017857

Add them up:

0.0021277 + 0.0030303 + 0.0017857 ≈ 0.0069437

Now take reciprocal to get R_T:

R_T = 1 / 0.0069437 ≈ 144.0 Ω

*(We’ll keep one decimal place for now — you can round later if needed.)*

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Step 2: Find total current (I_T)

Use Ohm’s Law: I_T = V_total / R_T

I_T = 9.8 V / 144.0 Ω ≈ 0.06806 A

Convert to milliamps (mA) since resistors are small:
0.06806 A × 1000 = 68.06 mA

We’ll round to 68.1 mA at the end.

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Step 3: Find current in each branch

Again, use Ohm’s Law for each resistor: I = V / R

Since voltage across each is 9.8 V:

- I₁ = 9.8 / 470 ≈ 0.02085 A = 20.85 mA
- I₂ = 9.8 / 330 ≈ 0.02970 A = 29.70 mA
- I₃ = 9.8 / 560 ≈ 0.01750 A = 17.50 mA

Check: Add them up → 20.85 + 29.70 + 17.50 = 68.05 mA — matches our total current! Good.

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Step 4: Power dissipated by each resistor

Power formula: P = V² / R (since we know voltage and resistance)

Or P = I × V — either works. Let’s use P = V² / R for consistency.

- P₁ = (9.8)² / 470 = 96.04 / 470 ≈ 0.2043 W = 204.3 mW
- P₂ = 96.04 / 330 ≈ 0.2910 W = 291.0 mW
- P₃ = 96.04 / 560 ≈ 0.1715 W = 171.5 mW

Total power = sum of all powers:

P_total = 204.3 + 291.0 + 171.5 = 666.8 mW

Alternatively, check with P_total = V × I_T = 9.8 V × 0.06806 A ≈ 0.667 W = 667 mW — close enough (rounding difference).

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Final Answers (rounded appropriately):

- Total Resistance (R_T): 144.0 Ω
- Total Current (I_T): 68.1 mA
- Branch Currents:
- I₁ = 20.9 mA
- I₂ = 29.7 mA
- I₃ = 17.5 mA
- Power Dissipated:
- P₁ = 204 mW
- P₂ = 291 mW
- P₃ = 172 mW
- P_total = 667 mW

*(Note: We rounded to 3 significant figures where appropriate, as typical for electronics problems.)*

Final Answer:
Total Resistance = 144.0 Ω
Total Current = 68.1 mA
Branch Currents: I₁ = 20.9 mA, I₂ = 29.7 mA, I₃ = 17.5 mA
Power Dissipated: P₁ = 204 mW, P₂ = 291 mW, P₃ = 172 mW, P_total = 667 mW