Parallel Series Circuits Lesson Plans & Worksheets - Free Printable

Let’s solve each problem one by one. We’ll use Ohm’s Law (V = I × R) and rules for series and parallel circuits.

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Problem 1: Find I_T in the circuit

We have three resistors in parallel:
R₁ = 2 Ω, R₂ = 5 Ω, R₃ = 6 Ω
Voltage source V_T = 9 V

In a parallel circuit, voltage across each resistor is the same as the source voltage → so each resistor has 9 V across it.

To find total current I_T, we can either:
- Find current through each resistor and add them up, OR
- Find equivalent resistance first, then use I_T = V_T / R_eq

Let’s do both to check.

Method 1: Add individual currents

I₁ = V / R₁ = 9 V / 2 Ω = 4.5 A
I₂ = 9 V / 5 Ω = 1.8 A
I₃ = 9 V / 6 Ω = 1.5 A

I_T = I₁ + I₂ + I₃ = 4.5 + 1.8 + 1.5 = 7.8 A

Method 2: Equivalent resistance

For parallel resistors:

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃
= 1/2 + 1/5 + 1/6

Find common denominator — let’s use 30:

= 15/30 + 6/30 + 5/30 = 26/30 = 13/15

So R_eq = 15/13 ≈ 1.1538 Ω

Then I_T = V_T / R_eq = 9 / (15/13) = 9 × (13/15) = 117/15 = 7.8 A

✔ Both methods give same answer.

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Problem 2: Complete the table for this circuit

Circuit: All resistors in series:
R₁ = 8 Ω, R₂ = 12 Ω, R₃ = 10 Ω
V_T = 12 V

In series:
- Current is same everywhere → I_T = I₁ = I₂ = I₃
- Total resistance R_T = R₁ + R₂ + R₃ = 8 + 12 + 10 = 30 Ω
- Total current I_T = V_T / R_T = 12 V / 30 Ω = 0.4 A

Now fill in voltages:

V₁ = I × R₁ = 0.4 A × 8 Ω = 3.2 V
V₂ = 0.4 A × 12 Ω = 4.8 V
V₃ = 0.4 A × 10 Ω = 4.0 V

Check: V₁ + V₂ + V₃ = 3.2 + 4.8 + 4.0 = 12 V → matches V_T ✔

Table filled:

| V_T = 12 V | V₁ = 3.2 V | V₂ = 4.8 V | V₃ = 4.0 V |
|------------|------------|------------|------------|
| R_T = 30 Ω | R₁ = 8 Ω | R₂ = 12 Ω | R₃ = 10 Ω |
| I_T = 0.4 A| I₁ = 0.4 A | I₂ = 0.4 A | I₃ = 0.4 A |

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Problem 3: Complete the table for this circuit

Circuit: Two resistors in parallel (R₁ = 25 Ω, R₂ = 30 Ω), connected in series with R₃ = 10 Ω
V_T = 10 V

Step 1: Find equivalent resistance of parallel part (R₁ and R₂)

1/R_parallel = 1/25 + 1/30

Common denominator = 150:

= 6/150 + 5/150 = 11/150
→ R_parallel = 150/11 ≈ 13.636 Ω

Step 2: Total resistance R_T = R_parallel + R₃ = 150/11 + 10 = 150/11 + 110/11 = 260/11 ≈ 23.636 Ω

Step 3: Total current I_T = V_T / R_T = 10 / (260/11) = 10 × 11 / 260 = 110 / 260 = 11/26 ≈ 0.4231 A

This current flows through R₃ and also into the parallel branch.

So:
I₃ = I_T = 11/26 A ≈ 0.4231 A
V₃ = I₃ × R = (11/26) × 10 = 110/26 = 55/13 ≈ 4.2308 V

Now, voltage across parallel branch (V_parallel) = V_T - V₃ = 10 - 55/13 = (130 - 55)/13 = 75/13 ≈ 5.7692 V

Since R₁ and R₂ are in parallel, they both have this voltage:

V₁ = V₂ = 75/13 V ≈ 5.7692 V

Now find currents:

I₁ = V₁ / R₁ = (75/13) / 25 = 75/(13×25) = 3/13 ≈ 0.2308 A
I₂ = V₂ / R₂ = (75/13) / 30 = 75/(13×30) = 5/26 ≈ 0.1923 A

Check: I₁ + I₂ = 3/13 + 5/26 = 6/26 + 5/26 = 11/26 = I_T ✔

Now fill table:

| V_T = 10 V | V₁ = 75/13 V ≈ 5.77 V | V₂ = 75/13 V ≈ 5.77 V | V₃ = 55/13 V ≈ 4.23 V |
|----------------|------------------------|------------------------|------------------------|
| R_T = 260/11 Ω ≈ 23.64 Ω | R₁ = 25 Ω | R₂ = 30 Ω | R₃ = 10 Ω |
| I_T = 11/26 A ≈ 0.423 A | I₁ = 3/13 A ≈ 0.231 A | I₂ = 5/26 A ≈ 0.192 A | I₃ = 11/26 A ≈ 0.423 A |

We can write exact fractions or decimals. Since the problem didn’t specify, we’ll use decimals rounded to 3 places for clarity.

Final values (rounded):

V₁ = 5.769 V → 5.77 V
V₂ = 5.769 V → 5.77 V
V₃ = 4.231 V → 4.23 V
R_T = 23.636 Ω → 23.64 Ω
I_T = 0.423 A
I₁ = 0.231 A
I₂ = 0.192 A
I₃ = 0.423 A

But let’s keep more precision if needed. Actually, since the inputs were whole numbers, maybe fractions are better? But tables usually expect decimals. Let’s go with 3 decimal places.

Wait — let me recalculate using decimals from start to avoid fraction confusion.

Alternative calculation with decimals:

R_parallel = 1 / (1/25 + 1/30) = 1 / (0.04 + 0.033333) = 1 / 0.073333 ≈ 13.6364 Ω
R_T = 13.6364 + 10 = 23.6364 Ω
I_T = 10 / 23.6364 ≈ 0.42308 A → 0.423 A

V₃ = I_T × R₃ = 0.42308 × 10 = 4.2308 V → 4.231 V

V_parallel = 10 - 4.2308 = 5.7692 V → 5.769 V

I₁ = 5.7692 / 25 = 0.23077 A → 0.231 A
I₂ = 5.7692 / 30 = 0.19231 A → 0.192 A

All checks out.

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Final Answer:

Problem 1:
I_T = 7.8 A

Problem 2 Table:

| V_T = 12 V | V₁ = 3.2 V | V₂ = 4.8 V | V₃ = 4.0 V |
|------------|------------|------------|------------|
| R_T = 30 Ω | R₁ = 8 Ω | R₂ = 12 Ω | R₃ = 10 Ω |
| I_T = 0.4 A| I₁ = 0.4 A | I₂ = 0.4 A | I₃ = 0.4 A |

Problem 3 Table:

| V_T = 10 V | V₁ = 5.77 V | V₂ = 5.77 V | V₃ = 4.23 V |
|----------------|-------------|-------------|-------------|
| R_T = 23.64 Ω | R₁ = 25 Ω | R₂ = 30 Ω | R₃ = 10 Ω |
| I_T = 0.423 A | I₁ = 0.231 A| I₂ = 0.192 A| I₃ = 0.423 A|

(Note: Values rounded to 3 significant figures where appropriate.)