Solved + Series Circuit Worksheet #4 Mects NATEF Task | Chegg.com - Free Printable

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

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🔧 Key Principles for Series Circuits:

1. Current is the same everywhere in a series circuit:
$$
I_T = I_1 = I_2 = I_3 = \dots
$$

2. Total resistance is the sum of all resistances:
$$
R_{\text{total}} = R_1 + R_2 + R_3 + \dots
$$

3. Voltage drops across each resistor:
$$
V = I \times R \quad \text{(Ohm's Law)}
$$

4. Power dissipated by a resistor:
$$
P = I^2 \times R \quad \text{or} \quad P = V \times I
$$

5. Total power:
$$
P_{\text{total}} = I_T^2 \times R_{\text{total}} \quad \text{or} \quad P_{\text{total}} = E \times I_T
$$

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We'll go through each problem one at a time.

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## ✔ Problem 1

Given:
- $ E = 6 $ volts
- $ I_T = 6 $ A
- $ R_1 = 1\ \Omega $
- $ R_2 = 1\ \Omega $
- $ R_3 = 1\ \Omega $
- $ R_4 = 1\ \Omega $

All resistors are in series, so:

Step 1: Total Resistance

$$
R_{\text{total}} = R_1 + R_2 + R_3 + R_4 = 1 + 1 + 1 + 1 = 4\ \Omega
$$

Step 2: Check if current matches Ohm’s Law

$$
I_T = \frac{E}{R_{\text{total}}} = \frac{6\ \text{V}}{4\ \Omega} = 1.5\ \text{A}
$$
But given $ I_T = 6\ \text{A} $? That contradicts the voltage and resistance!

Wait — this suggests an inconsistency. Let's double-check.

But the problem says: $ I_T = 6 $ A, $ E = 6 $ V → then total resistance must be:
$$
R_{\text{total}} = \frac{E}{I_T} = \frac{6}{6} = 1\ \Omega
$$

But we have four 1-ohm resistors in series → total = 4 Ω. Contradiction.

So either:
- The values are wrong, or
- We misread the diagram.

But looking again: all resistors are in series, so total resistance should be 4 Ω.

So if $ E = 6 $ V, $ R_{\text{total}} = 4\ \Omega $, then:
$$
I_T = \frac{6}{4} = 1.5\ \text{A}
$$

But the problem says $ I_T = 6 $ A — that can't be right unless the voltage is higher.

Wait — maybe there's a typo? Or perhaps the image has different values?

But let's assume the given values are correct, so maybe the voltage is not 6 V? But it says $ E = 6 $ V.

Alternatively, could the resistors not all be 1 Ω?

Wait — let's recheck:
It says:
- $ R_1 = 1\ \Omega $
- $ R_2 = 1\ \Omega $
- $ R_3 = 1\ \Omega $
- $ R_4 = 1\ \Omega $

Yes, all 1 Ω → total 4 Ω.

So with $ E = 6 $ V → $ I = 1.5 $ A.

But problem says $ I_T = 6 $ A → contradiction.

So likely, either the voltage is 24 V, or the current is 1.5 A.

But since both are given, we must use the given values, but they conflict.

Wait — maybe the current is not 6 A, but something else?

No — it says: “$ I_T = 6 $ amperes”

This is impossible with 4 Ω and 6 V.

So perhaps the diagram shows a different configuration, but from the drawing, it's clearly a series loop.

Alternatively, maybe the resistors are not all 1 ohm?

Wait — let's check again:

> R1 = 1 ohm
> R2 = 1 ohm
> R3 = 1 ohm
> R4 = 1 ohm

Yes.

So $ R_{\text{total}} = 4\ \Omega $

Then $ I_T = E / R_{\text{total}} = 6 / 4 = 1.5\ \text{A} $

But problem says $ I_T = 6 $ A → this is not possible unless the voltage is 24 V.

But it says $ E = 6 $ V.

So there's a mistake in the problem statement.

But wait — maybe the current is not 6 A, but we're to calculate it?

No — it says "I_T = 6 amperes" as given.

But that can't be.

Unless... is the circuit not purely series?

Looking at the diagram: It appears to be a single loop: battery → R1 → R2 → R3 → R4 → back to battery → series circuit.

So no parallel paths.

So either the values are inconsistent, or we’re supposed to ignore one.

But let’s suppose that the given $ I_T = 6 $ A is correct, then we can calculate E:

$$
E = I_T \times R_{\text{total}} = 6\ \text{A} \times 4\ \Omega = 24\ \text{V}
$$

But it says $ E = 6 $ V — contradiction.

Alternatively, suppose $ E = 6 $ V is correct, then $ I_T = 1.5 $ A.

But problem says $ I_T = 6 $ A.

So likely, the current is not 6 A — maybe it's a typo.

Wait — look at Problem 3: it says $ I_T = 2 $ A, $ E = 24 $ V → consistent.

So perhaps in Problem 1, the current is not 6 A, but we need to calculate it?

But it says: “I_T = 6 amperes” — so it's given.

This is problematic.

Wait — perhaps the resistors are not all 1 ohm?

Let me re-read:

> R1 = 1 ohm
> R2 = 1 ohm
> R3 = 1 ohm
> R4 = 1 ohm

Yes.

So unless the diagram is not series, but it looks like it is.

Wait — could R3 and R4 be in parallel? No — the diagram shows a single path.

So I think there is a typo in Problem 1.

But let’s assume the given values are correct, and proceed with the given current, even if it contradicts voltage.

But that would mean the voltage is not 6 V — but it says it is.

Alternatively, maybe the battery is 24 V, but written as 6 V?

No — it says E = 6 volts.

So let’s ignore the contradiction and use Ohm’s Law properly.

So let's recalculate based on correct physics.

Correct Approach for Problem 1:

Given:
- $ E = 6 $ V
- $ R_1 = R_2 = R_3 = R_4 = 1\ \Omega $
- So $ R_{\text{total}} = 4\ \Omega $

Then:
$$
I_T = \frac{E}{R_{\text{total}}} = \frac{6}{4} = 1.5\ \text{A}
$$

But problem says $ I_T = 6 $ A — which is wrong.

So likely, the given current is incorrect, or the voltage is 24 V.

But let's see Problem 3: $ E = 24 $ V, $ I_T = 2 $ A → $ R_{\text{total}} = 12\ \Omega $

So maybe in Problem 1, the current is not 6 A, but we're to calculate it.

But the blank says: “I_T = 6 amperes” — suggesting it's given.

This is confusing.

Wait — maybe the resistors are not all 1 ohm?

Look again:

> R1 = 1 ohm
> R2 = 1 ohm
> R3 = 1 ohm
> R4 = 1 ohm

Yes.

So unless the circuit is not series, but it is.

Perhaps the battery is 24 V, but labeled as 6 V? Unlikely.

Alternatively, maybe the current is 1.5 A, and the “6 amperes” is a typo.

But let’s assume that the current is indeed 6 A, then the voltage must be 24 V, but it says 6 V.

So this is inconsistent.

But let’s move to Problem 2, and see if it makes sense.

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## ✔ Problem 2

Given:
- $ E = 3 $ volts
- $ I_T = 3 $ amperes
- $ R_1 = 0.5\ \Omega $
- $ R_2 = 1\ \Omega $
- $ R_3 = 0.5\ \Omega $
- $ R_4 = 1\ \Omega $

Check consistency:
- $ R_{\text{total}} = 0.5 + 1 + 0.5 + 1 = 3\ \Omega $
- $ I_T = E / R_{\text{total}} = 3 / 3 = 1\ \text{A} $
- But given $ I_T = 3 $ A → again contradiction.

Same issue.

So either:
- $ I_T = 1 $ A, or
- $ E = 9 $ V

But given $ E = 3 $ V, $ I_T = 3 $ A → implies $ R_{\text{total}} = 1\ \Omega $, but actual is 3 Ω.

So again, inconsistent.

Wait — perhaps the current is not given? But it says “I_T = 3 amperes”.

Maybe the values are for different problems?

Let’s look at Problem 3:

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## ✔ Problem 3

Given:
- $ E = 24 $ V
- $ I_T = 2 $ A
- $ R_1 = 2\ \Omega $
- $ R_2 = 4\ \Omega $
- $ R_3 = ? $
- $ R_4 = 1\ \Omega $
- $ R_5 = 1\ \Omega $

First, find $ R_{\text{total}} $:
$$
R_{\text{total}} = \frac{E}{I_T} = \frac{24}{2} = 12\ \Omega
$$

Sum of known resistors:
$$
R_1 + R_2 + R_4 + R_5 = 2 + 4 + 1 + 1 = 8\ \Omega
$$

So $ R_3 = 12 - 8 = 4\ \Omega $

Now fill in the blanks.

Currents:

In series, all currents equal:
- $ I_1 = I_2 = I_3 = I_4 = I_5 = I_T = 2 $ A

So:
- $ I_1 = 2 $ A
- $ I_4 = 2 $ A

Voltage Drops:

Use $ V = I \times R $

- $ E_2 = I_2 \times R_2 = 2 \times 4 = 8 $ V
- $ E_5 = I_5 \times R_5 = 2 \times 1 = 2 $ V

Power:

- $ P_3 = I_3^2 \times R_3 = (2)^2 \times 4 = 4 \times 4 = 16 $ W
- $ P_5 = I_5^2 \times R_5 = 4 \times 1 = 4 $ W

Total Power:

- $ P_{\text{total}} = E \times I_T = 24 \times 2 = 48 $ W
or $ I_T^2 \times R_{\text{total}} = 4 \times 12 = 48 $ W

So answers for Problem 3:

- $ I_1 = 2 $ A
- $ I_4 = 2 $ A
- $ E_2 = 8 $ V
- $ E_5 = 2 $ V
- $ P_3 = 16 $ W
- $ P_5 = 4 $ W
- $ P_{\text{total}} = 48 $ W
- $ R_3 = 4\ \Omega $

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## ✔ Problem 4

Given:
- $ E = 12 $ V
- $ I_T = 1 $ A
- $ R_1 = 2\ \Omega $
- $ R_2 = ? $
- $ R_3 = 4\ \Omega $
- $ R_4 = 3\ \Omega $
- $ R_5 = 1\ \Omega $

Find $ R_{\text{total}} $:
$$
R_{\text{total}} = \frac{E}{I_T} = \frac{12}{1} = 12\ \Omega
$$

Sum of known resistors:
$$
R_1 + R_3 + R_4 + R_5 = 2 + 4 + 3 + 1 = 10\ \Omega
$$

So $ R_2 = 12 - 10 = 2\ \Omega $

Now fill in:

Currents:

Series → all currents = $ I_T = 1 $ A
- $ I_3 = 1 $ A
- $ I_5 = 1 $ A

Voltage Drops:

- $ E_1 = I_1 \times R_1 = 1 \times 2 = 2 $ V
- $ E_4 = I_4 \times R_4 = 1 \times 3 = 3 $ V

Powers:

- $ P_2 = I_2^2 \times R_2 = (1)^2 \times 2 = 2 $ W
- $ P_4 = I_4^2 \times R_4 = 1 \times 3 = 3 $ W

Total Power:

- $ P_{\text{total}} = E \times I_T = 12 \times 1 = 12 $ W

So answers for Problem 4:

- $ I_3 = 1 $ A
- $ I_5 = 1 $ A
- $ E_1 = 2 $ V
- $ E_4 = 3 $ V
- $ P_2 = 2 $ W
- $ P_4 = 3 $ W
- $ P_{\text{total}} = 12 $ W
- $ R_2 = 2\ \Omega $

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Now back to Problems 1 and 2 — they have inconsistent data.

Let’s re-express them assuming the current is not given, or the voltage is not given.

But the problem says:

> Problem 1: $ I_T = 6 $ A, $ E = 6 $ V → impossible with 4 Ω

Wait — unless the resistors are not all 1 ohm?

Wait — let’s check Problem 1 again:

> R1 = 1 ohm
> R2 = 1 ohm
> R3 = 1 ohm
> R4 = 1 ohm

Yes.

But if $ I_T = 6 $ A, then $ E = I_T \times R_{\text{total}} = 6 \times 4 = 24 $ V

But it says $ E = 6 $ V → contradiction.

So likely, the current is not 6 A, but rather we are to calculate it.

But the blank says: “I_T = 6 amperes” — suggesting it's given.

Alternatively, maybe the value of R3 or R4 is different?

Wait — in Problem 1, it says:

> R1 = 1 ohm
> R2 = 1 ohm
> R3 = 1 ohm
> R4 = 1 ohm

But perhaps R3 is not 1 ohm? No — it says so.

Wait — maybe the circuit is not series? But the diagram shows a single loop.

Another possibility: the battery is 24 V, but labeled as 6 V? Unlikely.

Or maybe the current is 1.5 A, and “6 amperes” is a typo.

Let’s assume that “I_T = 6 amperes” is a typo, and we should use $ E = 6 $ V, $ R_{\text{total}} = 4\ \Omega $ → $ I_T = 1.5 $ A

Then proceed.

Similarly for Problem 2.

But let’s try solving Problem 1 with corrected current.

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## ✔ Re-solving Problem 1 (Assuming typo in current)

Given:
- $ E = 6 $ V
- $ R_1 = R_2 = R_3 = R_4 = 1\ \Omega $
- So $ R_{\text{total}} = 4\ \Omega $
- $ I_T = \frac{6}{4} = 1.5 $ A

So:
- $ I_2 = I_T = 1.5 $ A

Voltage drops:
- $ E_2 = I_2 \times R_2 = 1.5 \times 1 = 1.5 $ V
- $ E_3 = I_3 \times R_3 = 1.5 \times 1 = 1.5 $ V

Powers:
- $ P_1 = I_1^2 \times R_1 = (1.5)^2 \times 1 = 2.25 $ W
- $ P_4 = (1.5)^2 \times 1 = 2.25 $ W

Total power:
- $ P_{\text{total}} = E \times I_T = 6 \times 1.5 = 9 $ W
or $ I_T^2 \times R_{\text{total}} = 2.25 \times 4 = 9 $ W

So answers:
- $ I_2 = 1.5 $ A
- $ E_2 = 1.5 $ V
- $ E_3 = 1.5 $ V
- $ P_1 = 2.25 $ W
- $ P_4 = 2.25 $ W
- $ P_{\text{total}} = 9 $ W

But problem says $ I_T = 6 $ A — so unless we accept that, we can’t use it.

Alternatively, maybe the resistors are not all 1 ohm?

Wait — look at Problem 2:

> R1 = 0.5 ohm
> R2 = 1 ohm
> R3 = 0.5 ohm
> R4 = 1 ohm
> Sum = 3 ohms
> E = 3 V → I_T = 1 A
> But given I_T = 3 A → still contradiction

So likely, in both Problems 1 and 2, the current is not given correctly.

But in Problem 3 and 4, the values are consistent.

So probably, in Problems 1 and 2, the current is to be calculated, not given.

But the worksheet says: “I_T = 6 amperes” — so it's presented as given.

Perhaps the voltage is not 6 V, but 24 V?

Let’s assume for Problem 1 that $ E = 24 $ V, $ I_T = 6 $ A → then $ R_{\text{total}} = 4\ \Omega $, which matches.

So maybe $ E = 24 $ V, but written as 6 V? Probably a typo.

Similarly, in Problem 2: if $ E = 9 $ V, $ I_T = 3 $ A → $ R_{\text{total}} = 3\ \Omega $, which matches.

So likely, the voltages are wrong.

But let’s assume the current is correct, and voltage is unknown, or vice versa.

But since the battery is labeled $ E $, and current is given, we’ll use that.

Let’s solve Problem 1 with $ I_T = 6 $ A, and $ R_{\text{total}} = 4\ \Omega $, then:

$$
E = I_T \times R_{\text{total}} = 6 \times 4 = 24\ \text{V}
$$

So $ E = 24 $ V (not 6 V) — so the given $ E = 6 $ V is wrong.

Similarly, for Problem 2: $ I_T = 3 $ A, $ R_{\text{total}} = 3\ \Omega $ → $ E = 9 $ V

But it says $ E = 3 $ V — also wrong.

So likely, the voltage values are typos.

Let’s proceed with corrected voltages based on given current and resistances.

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## ✔ Final Solutions (Correcting Voltage Typos)

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🔹 Problem 1

Assume $ I_T = 6 $ A is correct, $ R_{\text{total}} = 4\ \Omega $ → $ E = 24 $ V (not 6 V)

Then:

- $ I_2 = 6 $ A (same as IT)
- $ E_2 = I_2 \times R_2 = 6 \times 1 = 6 $ V
- $ E_3 = 6 \times 1 = 6 $ V
- $ P_1 = I_1^2 \times R_1 = 36 \times 1 = 36 $ W
- $ P_4 = 36 \times 1 = 36 $ W
- $ P_{\text{total}} = I_T^2 \times R_{\text{total}} = 36 \times 4 = 144 $ W

Or $ P_{\text{total}} = E \times I_T = 24 \times 6 = 144 $ W

So answers:
- $ I_2 = 6 $ A
- $ E_2 = 6 $ V
- $ E_3 = 6 $ V
- $ P_1 = 36 $ W
- $ P_4 = 36 $ W
- $ P_{\text{total}} = 144 $ W

(And $ E = 24 $ V, not 6 V)

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🔹 Problem 2

$ I_T = 3 $ A, $ R_{\text{total}} = 0.5+1+0.5+1 = 3\ \Omega $ → $ E = 3 \times 3 = 9 $ V

- $ I_1 = 3 $ A
- $ E_1 = 3 \times 0.5 = 1.5 $ V
- $ E_4 = 3 \times 1 = 3 $ V
- $ P_2 = (3)^2 \times 1 = 9 $ W
- $ P_3 = 9 \times 0.5 = 4.5 $ W
- $ P_{\text{total}} = 9 \times 3 = 27 $ W

So:
- $ I_1 = 3 $ A
- $ E_1 = 1.5 $ V
- $ E_4 = 3 $ V
- $ P_2 = 9 $ W
- $ P_3 = 4.5 $ W
- $ P_{\text{total}} = 27 $ W

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✔ Problem 3 (already solved)

- $ I_1 = 2 $ A
- $ I_4 = 2 $ A
- $ E_2 = 8 $ V
- $ E_5 = 2 $ V
- $ P_3 = 16 $ W
- $ P_5 = 4 $ W
- $ P_{\text{total}} = 48 $ W
- $ R_3 = 4\ \Omega $

---

✔ Problem 4 (already solved)

- $ I_3 = 1 $ A
- $ I_5 = 1 $ A
- $ E_1 = 2 $ V
- $ E_4 = 3 $ V
- $ P_2 = 2 $ W
- $ P_4 = 3 $ W
- $ P_{\text{total}} = 12 $ W
- $ R_2 = 2\ \Omega $

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## 🟩 Final Answer Table

| Problem | I2/I1 | I4/I5 | E2/E1 | E3/E4 | P1/P2 | P4/P3 | P_total |
|--------|-------|-------|--------|--------|--------|--------|----------|
| 1 | 6 A | 6 A | 6 V | 6 V | 36 W | 36 W | 144 W |
| 2 | 3 A | 3 A | 1.5 V | 3 V | 9 W | 4.5 W | 27 W |
| 3 | 2 A | 2 A | 8 V | 2 V | 16 W | 4 W | 48 W |
| 4 | 1 A | 1 A | 2 V | 3 V | 2 W | 3 W | 12 W |

> Note: Voltages in Problems 1 and 2 were corrected from 6 V and 3 V to 24 V and 9 V respectively, due to inconsistencies.

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If you believe the original voltages are correct, then the currents must be 1.5 A and 1 A respectively, and the "6 A" and "3 A" are typos.

But based on the given current values, the voltages must be 24 V and 9 V.

So final answers are as above.

Let me know if you want the version assuming the voltages are correct and currents are to be calculated.