Solved Equivalent Resistance Worksheet Name: 2.0 2 www 8.0 Ω ... - Free Printable

Let’s solve each problem one by one.

---

Problem 8:

Circuit has:
- A parallel section: top branch = 8Ω + 2Ω = 10Ω; bottom branch = 10Ω → so two 10Ω resistors in parallel.
- Then that parallel combo is in series with a 60Ω and a 5Ω resistor.
- Voltage source = 400V.

Step 1: Find equivalent resistance of the parallel part.

Two 10Ω resistors in parallel:

R_parallel = (10 × 10) / (10 + 10) = 100 / 20 = 5Ω

Step 2: Add the series resistors.

Total R = 5Ω (parallel) + 60Ω + 5Ω = 70Ω

→ So, equivalent resistance = 70Ω

Step 3: Total current using Ohm’s Law: I = V / R = 400V / 70Ω ≈ 5.714 A

We can leave it as a fraction: 400/70 = 40/7 ≈ 5.714 A

---

Problem 9:

Circuit:
- Battery = 12V
- Resistors: 2Ω in series with a parallel combo of 20Ω and (10Ω || 10Ω)

First, find the 10Ω || 10Ω:

R = (10×10)/(10+10) = 100/20 = 5Ω

Now, that 5Ω is in series with 20Ω? Wait — looking at diagram:

Actually, the 20Ω is in parallel with the combination of 10Ω and 10Ω (which are in series? Or parallel?).

Wait — re-examining the diagram description:

It says: “20Ω” and then below it “10Ω” and “10Ω” — likely the two 10Ω are in parallel, and that whole thing is in parallel with the 20Ω? Or is the 20Ω in series with the parallel 10Ωs?

Looking again: The diagram shows:

Battery → 2Ω → then splits into two branches:
- One branch: 20Ω
- Other branch: 10Ω and 10Ω in series? Or parallel?

The way it's drawn: “20Ω” on top, and below it “10Ω” and “10Ω” — probably meaning the two 10Ω are in parallel, and that parallel combo is in parallel with the 20Ω? That would make sense.

But let’s read carefully: In problem 9, the circuit is:

12V battery → 2Ω resistor → then a parallel section with:
- Top: 20Ω
- Bottom: 10Ω and 10Ω — but how are they connected? If they’re drawn side by side under the 20Ω, likely they are in parallel with each other AND with the 20Ω? Or is the 10Ω and 10Ω in series, and that series combo is in parallel with 20Ω?

Actually, standard interpretation: if it says “10Ω” and “10Ω” stacked vertically under “20Ω”, it usually means the two 10Ω are in parallel, and that parallel pair is in parallel with the 20Ω.

So: three resistors in parallel: 20Ω, 10Ω, 10Ω.

Wait — no, that doesn’t match typical diagrams. Let me think differently.

Perhaps: after the 2Ω, there is a node that splits into two paths:
- Path 1: 20Ω
- Path 2: 10Ω and 10Ω in series → so 20Ω total for that path.

Then both paths are in parallel.

That makes more sense. Because if the two 10Ω were in parallel, it would be 5Ω, and then in parallel with 20Ω → but let’s check the math either way.

Assume: after 2Ω, we have two branches in parallel:
- Branch A: 20Ω
- Branch B: 10Ω + 10Ω = 20Ω (in series)

So both branches are 20Ω → parallel equivalent = (20×20)/(20+20) = 400/40 = 10Ω

Then total resistance = 2Ω + 10Ω = 12Ω

Total current = V/R = 12V / 12Ω = 1A

This seems clean and likely correct.

If instead the two 10Ω were in parallel (5Ω) and that in parallel with 20Ω, then:

R_parallel = 1 / (1/20 + 1/5) = 1 / (1/20 + 4/20) = 1/(5/20) = 20/5 = 4Ω

Then total R = 2Ω + 4Ω = 6Ω → I = 12/6 = 2A

But which is it?

Looking back at the original image description: for problem 9, it says:

“2Ω” then a box with “20Ω” on top, and below it “10Ω” and “10Ω” — this typically means the two 10Ω are in parallel with each other, and that combination is in parallel with the 20Ω. But that would be unusual because usually if they are meant to be in series, they’d be drawn end-to-end.

However, in many worksheets, when they draw two resistors stacked vertically under another, it often means they are in parallel with each other and together in parallel with the top one.

But let’s look at problem 10 for pattern.

In problem 10: “2Ω” then a parallel section with “6Ω and 10Ω” on top (series), and “4Ω” on bottom — so clearly, the top branch is 6+10=16Ω, bottom is 4Ω, so parallel combo of 16Ω and 4Ω.

Similarly, in problem 9, likely: top branch is 20Ω, bottom branch is 10Ω and 10Ω in series → 20Ω, so parallel of two 20Ω → 10Ω, plus 2Ω = 12Ω.

I think that’s the intended design.

So I’ll go with:

Parallel section: two branches, each 20Ω → equivalent 10Ω

Total R = 2Ω + 10Ω = 12Ω

Total I = 12V / 12Ω = 1A

---

Problem 10:

Circuit:
- 2Ω in series with a parallel combo.
- Parallel combo: top branch = 6Ω + 10Ω = 16Ω; bottom branch = 4Ω

So, parallel equivalent of 16Ω and 4Ω:

R_parallel = (16 × 4) / (16 + 4) = 64 / 20 = 3.2Ω

Total resistance = 2Ω + 3.2Ω = 5.2Ω

Part b: Current through the 2Ω resistor when 120V applied.

Since 2Ω is in series with the rest, the total current flows through it.

I_total = V / R_total = 120V / 5.2Ω

Calculate: 120 ÷ 5.2 = 1200 ÷ 52 = 300 ÷ 13 ≈ 23.0769 A

Or exactly: 120 / 5.2 = 1200 / 52 = 300 / 13 A

---

Now, summarizing answers:

Problem 8:
a. Equivalent resistance = 70Ω
b. Total current = 400/70 = 40/7 ≈ 5.714 A

Problem 9:
a. Equivalent resistance = 12Ω
b. Total current = 1A

Problem 10:
a. Equivalent resistance = 5.2Ω
b. Current through 2Ω resistor = 120 / 5.2 = 300/13 ≈ 23.077 A

All calculations verified.

Final Answer:

Problem 8a: 70 Ω
Problem 8b: 40/7 A (or approximately 5.71 A)
Problem 9a: 12 Ω
Problem 9b: 1 A
Problem 10a: 5.2 Ω
Problem 10b: 300/13 A (or approximately 23.08 A)