Solved Meets NATEF Task: (A6-A-2) Diagnose Electrical | Chegg.com - Free Printable

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

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Circuit 1:

Given:
- E = 24 volts
- I_T = ? (we need to find total current)
- R1 = 6 ohms, R2 = 6 ohms → these are in series → R1+R2 = 12 ohms
- R3 = 6 ohms, R4 = 6 ohms → also in series → R3+R4 = 12 ohms
- R5 = 6 ohms → alone on its branch

So we have three branches in parallel:
- Branch 1: 12 ohms
- Branch 2: 12 ohms
- Branch 3: 6 ohms

Find equivalent resistance of parallel branches:

1/R_total = 1/12 + 1/12 + 1/6
= 1/12 + 1/12 + 2/12 = 4/12 = 1/3
→ R_total = 3 ohms

Now, I_T = E / R_total = 24 V / 3 Ω = 8 amperes

✔ Final Answer for Circuit 1: I_T = 8 A

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Circuit 2:

Given:
- I_T = 12 amperes
- R1 = 8 ohms
- R2 = 4 ohms, R4 = 4 ohms → in series → 8 ohms
- R3 = 4 ohms

Branches in parallel:
- Branch 1: R1 = 8 ohms
- Branch 2: R2+R4 = 8 ohms
- Branch 3: R3 = 4 ohms

Same as Circuit 1! Three branches: 8Ω, 8Ω, 4Ω

1/R_total = 1/8 + 1/8 + 1/4 = 1/8 + 1/8 + 2/8 = 4/8 = 1/2
→ R_total = 2 ohms

E = I_T × R_total = 12 A × 2 Ω = 24 volts

✔ Final Answer for Circuit 2: E = 24 V

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Circuit 3:

Given:
- E = 12 volts
- I_T = ?
- R1 = 4 ohms
- R2 = 4 ohms, R3 = 4 ohms → in parallel? Wait — look at diagram.

Actually, from the drawing:
- R2 and R3 are in parallel with each other.
- That combination is in series with R4.
- Then that whole group is in parallel with R1? No — wait, let's trace:

From battery positive:
→ splits into two paths:
- Path 1: through R1 only
- Path 2: goes through R2 and R3 (which are in parallel), then through R4

Wait — actually, looking again:

It looks like:
- R2 and R3 are in parallel → combine them first.
- Then that combo is in series with R4.
- Then that entire branch (R2||R3 + R4) is in parallel with R1.

Yes.

Step 1: R2 || R3 = (4×4)/(4+4) = 16/8 = 2 ohms

Step 2: Add R4 in series → 2 + 2 = 4 ohms (this is the right branch)

Step 3: Now this 4-ohm branch is in parallel with R1 (also 4 ohms)

So total resistance: two 4-ohm branches in parallel → 4||4 = 2 ohms

I_T = E / R_total = 12 V / 2 Ω = 6 amperes

✔ Final Answer for Circuit 3: I_T = 6 A

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Circuit 4:

Given:
- I_T = 6 amperes
- R1 = 8 ohms
- R2 = 4 ohms, R3 = 4 ohms → in series? Let’s see.

Diagram shows:
- Left branch: R1 alone
- Right branch: R2 and R3 in series → 4 + 4 = 8 ohms

So two branches in parallel: both 8 ohms

Total resistance: 8 || 8 = 4 ohms

E = I_T × R_total = 6 A × 4 Ω = 24 volts

✔ Final Answer for Circuit 4: E = 24 V

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Circuit 5:

Given:
- E = 12 volts
- I_T = ?
- R1 = 1 ohm
- R2 = 2 ohms, R3 = 2 ohms → in parallel → (2×2)/(2+2) = 4/4 = 1 ohm
- R4 = 1 ohm

Now, how are they connected?

From diagram:
- Battery → R1 → then splits to R2 and R3 (parallel) → then rejoins → then R4 → back to battery.

So: R1 + (R2||R3) + R4 all in series.

We already found R2||R3 = 1 ohm

Total R = R1 + 1 + R4 = 1 + 1 + 1 = 3 ohms

I_T = E / R = 12 V / 3 Ω = 4 amperes

✔ Final Answer for Circuit 5: I_T = 4 A

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Circuit 6:

Given:
- E = 12 volts
- I_T = 4 amperes
- R1 = ? (unknown)
- R2 = 4 ohms
- R3 = 4 ohms, R4 = 8 ohms, R5 = 4 ohms

First, analyze structure:

From diagram:
- After R1, it splits into two main branches:
- Left branch: R3 and R4 in parallel
- Right branch: R2 and R5 in series

Then those two big branches are in parallel with each other, and that whole thing is in series with R1.

Step 1: Find resistance of left branch (R3 || R4)

R3 = 4Ω, R4 = 8Ω → (4×8)/(4+8) = 32/12 = 8/3 ≈ 2.6667 ohms

Step 2: Right branch: R2 + R5 = 4 + 4 = 8 ohms

Step 3: These two branches are in parallel:

Left: 8/3 Ω, Right: 8 Ω

1/R_parallel = 1/(8/3) + 1/8 = 3/8 + 1/8 = 4/8 = 1/2
→ R_parallel = 2 ohms

Step 4: Total circuit resistance = R1 + R_parallel = R1 + 2

But we know E = 12 V, I_T = 4 A → so total R = E / I_T = 12 / 4 = 3 ohms

Therefore:

R1 + 2 = 3 → R1 = 1 ohm

✔ Final Answer for Circuit 6: R1 = 1 Ω

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Final Answers:

1. I_T = 8 A
2. E = 24 V
3. I_T = 6 A
4. E = 24 V
5. I_T = 4 A
6. R1 = 1 Ω