50+ series and parallel resistors worksheets for 10th Grade on ... - Free Printable

Let's go through each question one by one and solve them step-by-step.

---

Question 1:

The picture shows an electrical circuit. This circuit is a series circuit because:

Circuit Description:
Three light bulbs connected in a single loop with a battery — all components are in a single path.

Options:
A) It has 3 light bulbs.
B) It uses a single battery.
C) The electrical current is divided between the three light bulbs.
D) The same current flows through all three light bulbs.

✔ Correct Answer: D) The same current flows through all three light bulbs.

Explanation:
In a series circuit, components are connected end-to-end, forming a single path for current. This means:
- The current is the same at every point in the circuit.
- Current does not divide — it stays constant through all components.
- Having 3 bulbs or a single battery doesn’t define a series circuit (parallel circuits can also have multiple bulbs and one battery).
- Option C is false because current is not divided in a series circuit — that happens in parallel circuits.

So, D is correct.

---

Question 2:

Total resistance in this circuit is:

Given resistors in series:
- R₁ = 100 Ω
- R₂ = 300 Ω
- R₃ = 30 Ω

All in series, so total resistance:

$$
R_T = R_1 + R_2 + R_3 = 100 + 300 + 30 = 430\ \Omega
$$

But wait — the options are:
A) 450 ohms
B) 30 ohms
C) 400 ohms
D) 100 ohms

Hmm... 430 Ω is not listed. Let’s double-check the values.

Wait — maybe there’s a typo? Or perhaps I misread.

Looking again:
R₁ = 100 Ω, R₂ = 300 Ω, R₃ = 30 Ω → sum = 430 Ω

But none of the choices match exactly. However, option A is 450, which is close but not correct.

Wait — could it be that R₃ is 50 Ω instead of 30? No, it says 30 Ω.

Alternatively, maybe the image shows something different?

Wait — let me recheck:
R₁ = 100 Ω
R₂ = 300 Ω
R₃ = 30 Ω
Sum = 100 + 300 + 30 = 430 Ω

But 430 Ω isn't among the choices.

However, if R₃ were 50 Ω, then 100+300+50=450 → matches A.

But it clearly says 30 Ω.

Alternatively, could it be a typo in the question?

Wait — maybe R₂ is 200 Ω? No, it says 300.

Alternatively, perhaps the diagram shows something else?

But based on what’s written:
R₁ = 100 Ω, R₂ = 300 Ω, R₃ = 30 Ω → 430 Ω

But since 430 Ω isn’t an option, and 450 Ω is closest, perhaps there’s a mistake.

Wait — look at Question 3: It uses the same circuit, and asks for current.

Let’s use Question 3 to verify.

---

Question 3: Total current through this circuit is

Voltage = 9 V
Total resistance = ?
We need to calculate current using Ohm’s Law:

$$
I = \frac{V}{R}
$$

If R = 430 Ω → I = 9 / 430 ≈ 0.0209 A ≈ 0.021 A

Now check options:
A) 0.02 amps
B) 9 amps
C) 50 amps
D) 0.09 amps

0.02 amps is closest to 0.0209 A, so likely the answer is A)

So even though total resistance is 430 Ω, they expect us to choose A) 0.02 amps as current.

Therefore, total resistance must be ~450 Ω to get 0.02 A?

Let’s reverse-engineer:

If I = 0.02 A, V = 9 V → R = V/I = 9 / 0.02 = 450 Ω

Ah! So total resistance must be 450 Ω.

That suggests R₃ is 50 Ω, not 30 Ω.

But the diagram says 30 Ω.

Wait — perhaps R₂ is 320 Ω? Or maybe it's a typo.

Wait — look carefully at the diagram:

> R₁ = 100 Ω
> R₂ = 300 Ω
> R₃ = 30 Ω

Sum = 430 Ω → I = 9 / 430 ≈ 0.0209 A → rounds to 0.02 A

So A) 0.02 amps is acceptable.

But then Question 2 asks for total resistance.

If R = 430 Ω, but options are:
A) 450 ohms
B) 30 ohms
C) 400 ohms
D) 100 ohms

None is 430.

But closest is A) 450? Or is it possible that R₃ is 50 Ω?

Wait — perhaps the label is wrong.

Alternatively, maybe R₁ = 100, R₂ = 300, R₃ = 50 → sum = 450 Ω → then:

- Q2: A) 450 ohms ✔
- Q3: I = 9 / 450 = 0.02 A → A) 0.02 amps ✔

So probably R₃ is 50 Ω, not 30 Ω.

But the diagram says 30 Ω.

Wait — look again:
"R₃ 30 Ω"

Yes, it says 30 Ω.

But that leads to inconsistency.

Unless... is the voltage not 9 V?

No, it says 9 V.

Wait — maybe R₁ = 100, R₂ = 300, R₃ = 30 → sum = 430 Ω

Then:
- Q2: no correct choice? But we must pick best.

But Q3: I = 9 / 430 ≈ 0.0209 → still close to 0.02 A

So A) 0.02 amps is correct.

Thus, for Q2, although 430 Ω isn’t listed, 400 Ω is closest, but not exact.

But wait — maybe I miscalculated.

Wait — is it possible that R₁ = 100, R₂ = 300, R₃ = 30 → sum = 430

But maybe the circuit has only two resistors?

No, three are shown.

Alternatively, perhaps the 30 Ω is not part of the series?

No, all are in series.

Wait — maybe the label is R₃ = 50 Ω?

But it says 30 Ω.

Perhaps it's a typo in the quiz.

But given the data, best we can do is assume R₃ = 50 Ω, or accept that Q2 has no correct answer, but Q3 has A)

Alternatively, perhaps R₁ = 100, R₂ = 300, R₃ = 30 → sum = 430 Ω

Then Q2: none of the above — but since it’s multiple choice, maybe C) 400 ohms is intended?

But that’s not accurate.

Wait — perhaps R₂ is 270 Ω? Then 100+270+30=400 → C) 400 ohms

But it says 300.

This is confusing.

Wait — let’s look at Question 4.

---

Question 4: Voltage dropped across the 300 ohm resistor

Assuming:
- V_total = 9 V
- R_total = 100 + 300 + 30 = 430 Ω
- I = 9 / 430 ≈ 0.0209 A

Voltage across 300 Ω:
$$
V = I \times R = 0.0209 \times 300 ≈ 6.27\ V
$$

Closest option: A) 6 V

So A) 6 V is likely correct.

But if R_total were 450 Ω, I = 9 / 450 = 0.02 A → V = 0.02 × 300 = 6 V → perfect.

So again, if R₃ = 50 Ω, then R_total = 450 Ω → consistent with all answers.

So likely, the diagram has a typo: R₃ should be 50 Ω, not 30 Ω.

Or perhaps the label is wrong.

But given that all answers align if R₃ = 50 Ω, we’ll proceed with that assumption.

So:

Revised Values:

- R₁ = 100 Ω
- R₂ = 300 Ω
- R₃ = 50 Ω → makes total 450 Ω

(Though labeled as 30 Ω, probably typo)

---

Now Solve Each Question:

---

Q1: Series circuit because?

✔ D) The same current flows through all three light bulbs.

Answer: D

---

Q2: Total resistance

R_total = 100 + 300 + 50 = 450 Ω

Answer: A) 450 ohms

---

Q3: Total current

I = V / R = 9 V / 450 Ω = 0.02 A

Answer: A) 0.02 amps

---

Q4: Voltage across 300 Ω resistor

I = 0.02 A
V = I × R = 0.02 × 300 = 6 V

Answer: A) 6 V

---

Q5: What is the total resistance (RT) across this circuit?

New circuit:
- Battery: 9 V
- R₁ = 3 kΩ (which is 3000 Ω)
- R₂ = 10 kΩ
- They are in series (connected end-to-end)

So total resistance:

$$
R_T = 3\ \text{k}\Omega + 10\ \text{k}\Omega = 13\ \text{k}\Omega
$$

But the question says "What is the total resistance (RT) across this circuit?"

And options aren't shown — but likely:

A) 13 kΩ
B) 7 kΩ
C) 30 kΩ
D) 10 kΩ

But since options aren't visible, we just compute.

Answer: 13 kΩ

But wait — the diagram shows:
- R₁ = 3 kΩ
- R₂ = 10 kΩ
- Connected in series from node 1 to node 2

So yes, RT = 3 + 10 = 13 kΩ

So Answer: 13 kΩ

But since options aren’t shown, we can’t pick letter.

But based on context, likely A) 13 kΩ

---

✔ Final Answers:

1. D — Same current flows through all bulbs
2. A — 450 ohms *(assuming R₃ is 50 Ω)*
3. A — 0.02 amps
4. A — 6 V
5. 13 kΩ — Total resistance is 13 kΩ

---

⚠️ Note:

There appears to be a labeling error in the diagram where R₃ is labeled 30 Ω, but to make the math work (and match the multiple-choice options), it should be 50 Ω.

Alternatively, if R₃ is truly 30 Ω, then:
- R_total = 430 Ω → not in options
- I = 9 / 430 ≈ 0.0209 A → still close to 0.02 A
- V across 300 Ω = 0.0209 × 300 ≈ 6.27 V → still close to 6 V

So A) 0.02 A and A) 6 V are still acceptable approximations.

But Q2 would have no exact answer.

So likely, R₃ is meant to be 50 Ω.

---

✔ Final Answer Summary:

| Question | Answer |
|--------|--------|
| 1 | D |
| 2 | A (450 ohms) |
| 3 | A (0.02 amps) |
| 4 | A (6 V) |
| 5 | 13 kΩ |

Let me know if you want the explanation for Q5 with diagram details.