Solved Parallel Circuit Worksheet 12 Meets NATEF Tasks ( AA ... - Free Printable

Let's solve each of the parallel circuit problems step by step, using the principles of parallel circuits and Ohm’s Law.

---

🔧 Key Concepts for Parallel Circuits:

1. Voltage across each branch is the same as the source voltage:
$$
V_{\text{branch}} = E
$$

2. Total current (Iₜ) is the sum of currents through each branch:
$$
I_T = I_1 + I_2 + I_3 + \dots
$$

3. Total resistance (Rₜ) in parallel:
$$
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots
$$

4. Ohm’s Law:
$$
I = \frac{V}{R}, \quad V = IR, \quad R = \frac{V}{I}
$$

---

Let’s go one by one:

---

Problem 1

Given:
- E = 12 volts
- R₁ = 4 Ω
- R₂ = 12 Ω
- R₃ = 4 Ω
- R₄ = 12 Ω

Find: Iₜ = ?

#### Step 1: Find current in each branch

All resistors are in parallel → voltage across each = 12 V

- I₁ = V / R₁ = 12 / 4 = 3 A
- I₂ = 12 / 12 = 1 A
- I₃ = 12 / 4 = 3 A
- I₄ = 12 / 12 = 1 A

#### Step 2: Total current
$$
I_T = 3 + 1 + 3 + 1 = \boxed{8 \text{ amperes}}
$$

✔ Answer: Iₜ = 8 A

---

Problem 2

Given:
- R₁ = 12 Ω
- R₂ = 12 Ω
- R₃ = 12 Ω
- Iₜ = 4 amperes
- R₄ = 12 Ω

Wait — R₄ is shown, but not listed in given values? Let's look again.

Actually, it says:
> R1 = 12 ohms
> R2 = 12 ohms
> R3 = 12 ohms
> R4 = 12 ohms
> Iₜ = 4 amperes

So all four resistors are 12 Ω in parallel.

We need to find E = ?

#### Step 1: Find total resistance

$$
\frac{1}{R_T} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}
\Rightarrow R_T = 3\ \Omega
$$

#### Step 2: Use Ohm’s Law
$$
E = I_T \times R_T = 4\ \text{A} \times 3\ \Omega = \boxed{12\ \text{volts}}
$$

✔ Answer: E = 12 volts

---

Problem 3

Given:
- E = ?
- Iₜ = 1 ampere
- R₁ = 2 Ω
- R₂ = 4 Ω
- R₃ = 6 Ω
- R₄ = 12 Ω

Need to find E = ?

#### Step 1: Find total resistance

$$
\frac{1}{R_T} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{12}
$$

Find common denominator (LCM of 2,4,6,12 = 12):

$$
= \frac{6}{12} + \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{12}{12} = 1
\Rightarrow R_T = 1\ \Omega
$$

#### Step 2: Use Ohm’s Law
$$
E = I_T \times R_T = 1\ \text{A} \times 1\ \Omega = \boxed{1\ \text{volt}}
$$

✔ Answer: E = 1 volt

---

Problem 4

Given:
- E = 12 volts
- R₁ = 8 Ω
- R₂ = 8 Ω
- R₃ = 4 Ω

Find: Iₜ = ?

#### Step 1: Find current in each branch

- I₁ = 12 / 8 = 1.5 A
- I₂ = 12 / 8 = 1.5 A
- I₃ = 12 / 4 = 3 A

#### Step 2: Total current
$$
I_T = 1.5 + 1.5 + 3 = \boxed{6\ \text{amperes}}
$$

✔ Answer: Iₜ = 6 A

---

Problem 5

Given:
- E = 12 volts
- Iₜ = 4 amperes
- R₁ = 12 Ω
- R₂ = 12 Ω
- R₃ = ?

Find: R₃ = ?

#### Step 1: Find current through R₁ and R₂

Since all are in parallel, voltage across each is 12 V.

- I₁ = 12 / 12 = 1 A
- I₂ = 12 / 12 = 1 A

#### Step 2: Current through R₃

Total current = 4 A
So:
$$
I_3 = I_T - I_1 - I_2 = 4 - 1 - 1 = 2\ \text{A}
$$

#### Step 3: Use Ohm’s Law to find R₃
$$
R_3 = \frac{V}{I_3} = \frac{12}{2} = \boxed{6\ \Omega}
$$

✔ Answer: R₃ = 6 Ω

---

Problem 6

Given:
- E = ?
- Iₜ = 2 amperes
- R₁ = ?
- R₂ = 24 Ω
- R₃ = 24 Ω

Find: E = ?, R₁ = ?

Wait — we have two unknowns: E and R₁. But we can find E first.

But wait — only Iₜ = 2 A, and R₂ and R₃ known.

Let’s see what we can do.

We don’t know R₁, so we can't directly compute total resistance unless we express it.

But perhaps we can assume that all resistors are in parallel, and use the fact that voltage is same across all branches.

Let’s suppose E is unknown, and R₁ is unknown.

We can write:

$$
I_T = I_1 + I_2 + I_3
$$

Where:
- I₂ = E / R₂ = E / 24
- I₃ = E / 24
- I₁ = E / R₁

So:
$$
I_T = \frac{E}{R_1} + \frac{E}{24} + \frac{E}{24} = \frac{E}{R_1} + \frac{2E}{24} = \frac{E}{R_1} + \frac{E}{12}
$$

We are told I_T = 2 A

So:
$$
\frac{E}{R_1} + \frac{E}{12} = 2
$$

But we have two variables, so we cannot solve unless more info is given.

Wait — look at the diagram: Is R₁ alone on one branch? Yes.

But maybe we’re missing something.

Wait — perhaps the question wants us to find E and R₁, but we need another condition.

Wait — look again at the problem:

It says:
> E = ?
> Iₜ = 2 amperes
> R₂ = 24 Ω
> R₃ = 24 Ω
> R₁ = ?

But no other data.

Hmm.

But here’s a possibility: Maybe R₁ is meant to be calculated if we assume symmetry or something?

Alternatively, perhaps there's a typo — maybe E is given elsewhere?

Wait — no, in this problem, only Iₜ = 2 A, R₂ = 24 Ω, R₃ = 24 Ω, and R₁ = ?

But we cannot determine both E and R₁ from just I_T.

Unless… maybe R₁ is also 24 Ω? But that would be an assumption.

Wait — perhaps we're supposed to find E, and R₁ in terms of E?

But no — the problem likely expects numerical answers.

Wait — let’s check if total current is 2 A, and two resistors are 24 Ω each.

Let’s suppose that R₁ is also 24 Ω, then:

- I₂ = E / 24
- I₃ = E / 24
- I₁ = E / 24

Then:
$$
I_T = 3 \times \frac{E}{24} = \frac{E}{8}
$$

Set equal to 2 A:
$$
\frac{E}{8} = 2 \Rightarrow E = 16\ \text{volts}
$$

Then R₁ = 24 Ω

But is that valid? Only if R₁ is also 24 Ω — but it's not stated.

Alternatively, maybe the problem assumes R₁ is unknown, but we can still solve if we realize that the total current depends on E and R₁.

But without knowing either E or R₁, we cannot solve.

Wait — perhaps E is the same as in previous problems? No, not necessarily.

Wait — maybe I misread.

Let me re-express:

We are given:
- I_T = 2 A
- R₂ = 24 Ω
- R₃ = 24 Ω
- R₁ = ?

And we need to find E and R₁?

But with only one equation, we can't solve.

Wait — unless R₁ is missing, but perhaps the diagram shows only three branches, and we are to find E and R₁?

But still — two unknowns.

Wait — perhaps E is given? No, it's blank.

Wait — maybe the answer is that we cannot determine E and R₁ uniquely?

But that seems unlikely.

Wait — perhaps R₁ is shorted or something? No.

Wait — let’s try to think differently.

Is it possible that R₁ is the only unknown, and E is to be found?

But still, we need another relation.

Wait — unless the total resistance can be expressed, but we don’t know R₁.

Wait — perhaps there's a typo, and I_T = 2 A, and R₂ = R₃ = 24 Ω, and we are to find E and R₁?

But we can’t.

Wait — unless R₁ is such that the total current is 2 A, and we need to find E and R₁.

But infinite solutions exist.

For example:

Suppose E = 12 V

Then:
- I₂ = 12 / 24 = 0.5 A
- I₃ = 0.5 A
- So I₁ = 2 - 0.5 - 0.5 = 1 A

Then R₁ = E / I₁ = 12 / 1 = 12 Ω

So E = 12 V, R₁ = 12 Ω

Another possibility: E = 24 V

Then:
- I₂ = 24 / 24 = 1 A
- I₃ = 1 A
- So I₁ = 0 A → R₁ = ∞ → open circuit → not possible

Or E = 6 V

- I₂ = 6 / 24 = 0.25 A
- I₃ = 0.25 A
- I₁ = 2 - 0.5 = 1.5 A
- R₁ = 6 / 1.5 = 4 Ω

So multiple solutions.

Thus, without additional information, we cannot determine E and R₁ uniquely.

But wait — maybe the diagram shows R₁ is connected differently?

Looking back at the image description: It's a parallel circuit with three branches: R₁, R₂, R₃.

No indication of any other component.

So unless E is implied to be known, or R₁ is known, we can't solve.

But in the worksheet, E is blank, R₁ is blank, and I_T = 2 A, R₂ = 24, R₃ = 24.

This suggests that perhaps R₁ is intended to be the same as R₂ and R₃, i.e., 24 Ω?

Let’s assume R₁ = 24 Ω, then:

- I₁ = E / 24
- I₂ = E / 24
- I₃ = E / 24
- I_T = 3 × (E / 24) = E / 8

Set equal to 2 A:
$$
E / 8 = 2 \Rightarrow E = 16\ \text{volts}
$$

Then R₁ = 24 Ω

So:
- E = 16 V
- R₁ = 24 Ω

This is a reasonable assumption if the circuit is symmetric.

Alternatively, maybe R₁ is different, but the problem expects us to find E in terms of R₁?

But since it asks for E = ? and R₁ = ?, and both are blank, likely the intention is that R₁ is also 24 Ω, and we solve accordingly.

So let’s go with that.

✔ Answer (assuming R₁ = 24 Ω):
- E = 16 volts
- R₁ = 24 Ω

But wait — is there another way?

Wait — perhaps R₁ is not 24 Ω, but the total current is 2 A, and R₂ and R₃ are 24 Ω, and we are to find E and R₁?

Still underdetermined.

But notice: In Problem 5, we had similar setup, but E was given.

Here, E is not given.

Wait — perhaps E is the same as in Problem 1–5? Like 12 V?

Try E = 12 V

Then:
- I₂ = 12 / 24 = 0.5 A
- I₃ = 0.5 A
- So I₁ = 2 - 0.5 - 0.5 = 1 A
- R₁ = 12 / 1 = 12 Ω

So E = 12 V, R₁ = 12 Ω

But why choose 12 V?

No justification.

Alternatively, perhaps R₁ is missing, and we are to find E and R₁ such that the total current is 2 A.

But without more constraints, we can't.

Wait — perhaps the problem is incomplete, or there's a typo.

But looking back — in the image, it says:

> E = ?
> I_T = 2 amperes
> R2 = 24 ohms
> R3 = 24 ohms
> R1 = ?

But no other data.

So unless R1 is known, or E is known, we can't solve.

Wait — perhaps the diagram shows R1 is the only one with unknown value, and others are known, but E is also unknown.

But still.

Wait — maybe E is the same as in Problem 1? But that's arbitrary.

Alternatively, perhaps the total resistance is known? No.

Wait — unless the circuit has only R2 and R3, and R1 is a typo?

But it's drawn.

Given the ambiguity, and since R2 and R3 are both 24 Ω, and I_T = 2 A, and if we assume R1 is also 24 Ω, then:

- Each branch: I = E / 24
- Total I = 3 × (E / 24) = E / 8 = 2 → E = 16 V

So:

✔ E = 16 volts, R₁ = 24 ohms

This is the most logical assumption.

So we'll go with that.

---

✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ I_T = 8 $ A |
| 2 | $ E = 12 $ V |
| 3 | $ E = 1 $ V |
| 4 | $ I_T = 6 $ A |
| 5 | $ R_3 = 6 $ Ω |
| 6 | $ E = 16 $ V, $ R_1 = 24 $ Ω |

---

📝 Summary of Solutions:

1. Iₜ = 8 A
2. E = 12 V
3. E = 1 V
4. Iₜ = 6 A
5. R₃ = 6 Ω
6. E = 16 V, R₁ = 24 Ω *(assumed symmetric)*

Let me know if you want a detailed diagram explanation or verification!