Electric Circuits Worksheet with Problems and Solutions

You’ve uploaded a worksheet titled “9.05 Electric Circuits Worksheet” with three problems involving equivalent resistance and circuit analysis. I’ll solve each problem step-by-step, correct any errors in the provided answers, and explain the reasoning clearly.

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## ✔ Problem 1: Determine the equivalent (total) resistance for each circuit

a) Resistors in Parallel: 7Ω, 5Ω, 2Ω

When resistors are connected in parallel, the equivalent resistance is calculated using:

\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]

Plug in the values:

\[
\frac{1}{R_{eq}} = \frac{1}{7} + \frac{1}{5} + \frac{1}{2}
\]

Find a common denominator (LCM of 7, 5, 2 is 70):

\[
\frac{1}{R_{eq}} = \frac{10}{70} + \frac{14}{70} + \frac{35}{70} = \frac{59}{70}
\]

So,

\[
R_{eq} = \frac{70}{59} \approx 1.186 \, \Omega
\]

✔ Correct Answer: \( \boxed{1.19 \, \Omega} \) (rounded to 3 significant figures)

> ❗ The student wrote “1/Req = 1/7 = 1/5 + 1/2”, which is incorrectly written — it should be 1/Req = 1/7 + 1/5 + 1/2. But the final calculation (70/59 ≈ 1.18 Ω) is numerically correct. We’ll round to 3 sig figs → 1.19 Ω.

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b) Resistors in Series: 2Ω and 5Ω

When resistors are in series, simply add them:

\[
R_{eq} = R_1 + R_2 = 2 + 5 = 7 \, \Omega
\]

✔ Correct Answer: \( \boxed{7.0 \, \Omega} \) (to 2 significant figures, since inputs are whole numbers)

> ✔️ Student’s answer is correct.

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c) Resistors in Series: 2Ω, 5Ω, 7Ω

Again, all in series → add them:

\[
R_{eq} = 2 + 5 + 7 = 14 \, \Omega
\]

✔ Correct Answer: \( \boxed{14 \, \Omega} \)

> ✔️ Student’s answer is correct.

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## ✔ Problem 2: Fill out the table for the parallel circuit

Circuit: Three resistors (R₁=10Ω, R₂=20Ω, R₃=30Ω) connected in parallel to a 6V battery.

In parallel circuits:
- Voltage across each resistor is the same as the source voltage → 6V
- Current through each resistor: \( I = \frac{V}{R} \)
- Total current = sum of individual currents
- Equivalent resistance: \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)

Let’s verify the table:

| Circuit Position | Voltage (V) | Current (A) | Resistance (Ω) |
|------------------|-------------|-------------|----------------|
| 1 | 6 | 0.6 | 10.0 |
| 2 | 6 | 0.3 | 20.0 |
| 3 | 6 | 0.2 | 30.0 |
| Total | 6.00 | 1.1 | Req = 60/11 ≈ 5.45 Ω |

✔️ All values are correct:

- I₁ = 6/10 = 0.6 A
- I₂ = 6/20 = 0.3 A
- I₃ = 6/30 = 0.2 A
- I_total = 0.6 + 0.3 + 0.2 = 1.1 A
- R_eq = V / I_total = 6 / 1.1 ≈ 5.45 Ω

Also, you can compute R_eq directly:

\[
\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{20} + \frac{1}{30} = \frac{6}{60} + \frac{3}{60} + \frac{2}{60} = \frac{11}{60} \Rightarrow R_{eq} = \frac{60}{11} \approx 5.45 \, \Omega
\]

✔ All values in the table are correct.

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## ✔ Problem 3: Fill out the table for the combination circuit

Circuit: R₁ (10Ω) in series with a parallel combination of R₂ (20Ω) and R₃ (30Ω).

This is a series-parallel circuit.

Step 1: Find equivalent resistance of the parallel part (R₂ || R₃)

\[
\frac{1}{R_{23}} = \frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \Rightarrow R_{23} = 12 \, \Omega
\]

Step 2: Total resistance of circuit

R₁ (10Ω) in series with R₂₃ (12Ω):

\[
R_{total} = 10 + 12 = 22 \, \Omega
\]

Step 3: Total current from battery (assuming battery voltage is not given, but we can deduce it from data)

From the table:

- Position 1: V=2.7V, I=0.27A → R=10Ω → checks out: 2.7 / 0.27 = 10Ω
- Position 2: V=3.3V, I=0.165A → R=20Ω → 3.3 / 0.165 = 20Ω ✔️
- Position 3: V=3.3V, I=0.11A → R=30Ω → 3.3 / 0.11 = 30Ω ✔️

Note: R₂ and R₃ are in parallel → they have same voltage → 3.3V ✔️

Current through R₁ (the series resistor) must equal total current → 0.27A

Voltage across R₁ = I × R₁ = 0.27 × 10 = 2.7V

Voltage across parallel branch = 3.3V

Total battery voltage = 2.7V + 3.3V = 6.0V

Now check total current:

I_total = 0.27A (through R₁)

Current through R₂ = 0.165A

Current through R₃ = 0.11A

Sum: 0.165 + 0.11 = 0.275A — wait! This does not match 0.27A.

There’s a small rounding error here.

Let’s recalculate precisely:

If battery = 6V, R_total = 22Ω → I_total = 6 / 22 ≈ 0.2727 A

Then:

- Voltage across R₁ = I × R₁ = 0.2727 × 10 ≈ 2.727 V
- Voltage across parallel branch = 6 - 2.727 ≈ 3.273 V

Then:

- I₂ = 3.273 / 20 ≈ 0.1636 A
- I₃ = 3.273 / 30 ≈ 0.1091 A

Sum: 0.1636 + 0.1091 ≈ 0.2727 A → matches total current.

But the table uses rounded values:

- V₁ = 2.7V → implies I = 0.27A
- V₂ = V₃ = 3.3V → implies I₂ = 3.3/20 = 0.165A, I₃ = 3.3/30 = 0.11A → sum = 0.275A ≠ 0.27A

So there’s an inconsistency due to rounding.

However, since the problem says “fill out the table” and provides these values, we assume they are acceptable approximations for educational purposes.

✔ The table as filled is acceptable for this level, though technically the currents don’t perfectly add up due to rounding.

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## 📝 Final Summary of Answers:

Problem 1:

a) \( \boxed{1.19 \, \Omega} \)
b) \( \boxed{7.0 \, \Omega} \)
c) \( \boxed{14 \, \Omega} \)

Problem 2:

Table is correct as filled.
Explanation: All resistors in parallel → same voltage; currents calculated via Ohm’s Law; total current and Req verified.

Problem 3:

Table is approximately correct — slight rounding discrepancy in current sum (0.275A vs 0.27A), but acceptable for worksheet context.

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## 🎯 Tips for Improvement:

- In Problem 1a, write the formula correctly:
\( \frac{1}{R_{eq}} = \frac{1}{7} + \frac{1}{5} + \frac{1}{2} \), not “1/7 = 1/5 + 1/2”
- In Problem 3, if you’re calculating from scratch, use more precise decimals or keep fractions to avoid rounding errors.
- Always check that total current in series equals sum of branch currents in parallel sections.

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✔ You’ve done well overall! Just fix the notation in 1a and be mindful of rounding in combination circuits.

Let me know if you’d like to simulate this in PhET or need help drawing the circuits!