Solved PREVIEW WORKSHEET FOR EXPERIMENT C RESISTANCE IN | Chegg.com - Free Printable

Let’s solve each circuit step by step. We’ll find the equivalent resistance between points A and B for each diagram.

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a. Three resistors in series: 20 Ω, 40 Ω, and... wait — looking at the diagram, it shows only two labeled resistors? But the answer written is 60 Ω. Let me check again.

Actually, from left to right: first resistor is unlabeled but likely 20 Ω (as per common problems), then 40 Ω. Wait — no, the diagram says “20 Ω” on the first one? Actually, re-examining: the first resistor has “20 Ω” above it? No — in the image, part a shows:

A — [resistor] — [40 Ω] — B
And below the first resistor it says “20 Ω”? Actually, in many such worksheets, if not labeled, we assume based on context. But here, since the answer given is 60 Ω, and there are two resistors shown, probably they are 20 Ω and 40 Ω in series.

So:
R_eq = 20 + 40 = 60 Ω

✔ Correct.

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b. Resistors in series: (8+1) Ω, (13+1) Ω, (5+1) Ω → that’s 9 Ω, 14 Ω, 6 Ω.

Add them: 9 + 14 + 6 = 29 Ω

But the handwritten answer says “(26 ± 1.72) Ω” — that seems wrong. Maybe misread?

Wait — perhaps the labels are meant to be added as written: (8+1)=9, (13+1)=14, (5+1)=6 → total 29 Ω.

Unless there’s a typo or mislabeling — but based on what’s written, it should be 29 Ω.

However, let’s double-check: maybe the “(8+1)” means something else? Unlikely. Probably just 9, 14, 6 in series → 29 Ω.

But the student wrote 26 ± 1.72 — which suggests uncertainty or error. Since this is a preview worksheet, maybe they’re supposed to calculate nominal value.

We’ll go with 29 Ω.

Wait — hold on! Looking again: the diagram might have three resistors labeled as (8+1), (13+1), (5+1). So yes, 9 + 14 + 6 = 29 Ω.

But why did the student write 26? Maybe miscalculation: 8+13+5=26, forgetting the +1s? That would be incorrect.

Correct answer: 29 Ω

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c. Two resistors in parallel: 6 Ω and 3 Ω.

Formula for parallel:
1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2
→ R_eq = 2 Ω

✔ Matches handwritten answer.

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d. Three resistors in parallel: 10 Ω, 40 Ω, 25 Ω.

1/R_eq = 1/10 + 1/40 + 1/25

Find common denominator. LCM of 10, 40, 25.

10 = 2×5
40 = 2³×5
25 = 5²
LCM = 2³ × 5² = 8 × 25 = 200

Convert:

1/10 = 20/200
1/40 = 5/200
1/25 = 8/200

Sum: 20 + 5 + 8 = 33/200

So R_eq = 200 / 33 ≈ 6.0606... Ω → rounds to 6.06 Ω

✔ Matches handwritten answer.

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e. First, 5 Ω in series with a parallel combination of 12 Ω and 6 Ω.

First, find parallel part:
1/R_parallel = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4 → R_parallel = 4 Ω

Then add series 5 Ω: 5 + 4 = 9 Ω

✔ Matches handwritten answer.

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f. This one: 2 Ω in series with a parallel group of three 9 Ω resistors, then another 6 Ω in series.

First, parallel group: three 9 Ω resistors in parallel.

1/R_parallel = 1/9 + 1/9 + 1/9 = 3/9 = 1/3 → R_parallel = 3 Ω

Now total: 2 Ω + 3 Ω + 6 Ω = 11 Ω

Handwritten answer says “11 Ω” — correct.

Wait — user wrote “R = 11 Ω” — yes.

✔ Correct.

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g. Left side: 9 Ω in series with a parallel combination.

The parallel part has two branches:

Top branch: 4 Ω and 5 Ω in series → 4 + 5 = 9 Ω
Bottom branch: 18 Ω

So now we have two resistors in parallel: 9 Ω and 18 Ω.

1/R_parallel = 1/9 + 1/18 = 2/18 + 1/18 = 3/18 = 1/6 → R_parallel = 6 Ω

Now add the initial 9 Ω in series: 9 + 6 = 15 Ω

✔ Matches handwritten answer.

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Now, summarizing all answers correctly:

a. 60 Ω
b. 29 Ω (not 26 — student made mistake)
c. 2 Ω
d. 6.06 Ω
e. 9 Ω
f. 11 Ω
g. 15 Ω

But note: for part b, if the problem intended something else, but based on labeling, 29 Ω is correct.

However, let me recheck part b once more.

Diagram b: A — [(8+1)Ω] — [(13+1)Ω] — [(5+1)Ω] — B

That’s clearly 9, 14, 6 → sum 29.

Student wrote “(26 ± 1.72)” — perhaps they thought 8+13+5=26 and ignored the +1s? Or maybe measurement uncertainty? But since it's a theoretical problem, we take exact values.

So final answers:

a. 60 Ω
b. 29 Ω
c. 2 Ω
d. 6.06 Ω
e. 9 Ω
f. 11 Ω
g. 15 Ω

Final Answer:
a. 60 Ω
b. 29 Ω
c. 2 Ω
d. 6.06 Ω
e. 9 Ω
f. 11 Ω
g. 15 Ω