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 given $ I_T = 6 $ A → inconsistent.

So likely, the value of $ I_T $ is incorrect, or $ E $ is wrong, or resistors are not all 1 Ω.

But since this is a worksheet, maybe we are supposed to use the given current and calculate everything accordingly, even if it contradicts Ohm's law?

But that would be invalid.

Alternatively, perhaps the 6 A is not the total current, but no — it says $ I_T = 6 $ A.

Wait — unless the voltage is not 6 V, but something else?

No — it says $ E = 6 $ volts.

This is a contradiction.

But let's suppose that the given $ I_T = 6 $ A is correct, and $ E = 6 $ V, then total resistance must be:
$$
R_{\text{total}} = E / I_T = 6 / 6 = 1\ \Omega
$$

But sum of resistors is 4 Ω → impossible.

So unless the resistors are not all in series — but the diagram shows a single loop.

Wait — maybe R3 and R4 are not in series? No — the diagram shows a single path.

Perhaps the value of R3 is not 1 ohm? But it says R3 = 1 ohm.

Wait — let's look at the actual text:

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

Yes.

So this problem has conflicting data.

But let's assume the values are correct, and proceed with calculating based on Ohm’s Law.

So ignore the $ I_T = 6 $ A — because it doesn’t match.

Instead, use:
- $ E = 6 $ V
- $ R_{\text{total}} = 4\ \Omega $
- So $ I_T = 6 / 4 = 1.5 $ A

But the problem says $ I_T = 6 $ A — so we have a problem.

Alternatively, perhaps the resistors are not all 1 ohm?

Wait — maybe R3 = 0.5 ohms or something?

No — it says R3 = 1 ohm.

Wait — let's try to assume the current is correct, and see what happens.

Suppose $ I_T = 6 $ A, and $ E = 6 $ V → then $ R_{\text{total}} = 1\ \Omega $

But we have 4 resistors each 1 Ω → 4 Ω → contradiction.

So there must be a typo.

But let's check other problems.

Problem 2:
- $ E = 3 $ V
- $ I_T = 3 $ A → then $ R_{\text{total}} = 1\ \Omega $
- Resistors: R1=0.5, R2=1, R3=0.5, R4=1 → sum = 3 Ω → again, contradiction.

Same issue.

So maybe the current is not 6 A, but rather the voltage is 24 V?

Wait — in problem 3: $ E = 24 $ V, $ I_T = 2 $ A → $ R_{\text{total}} = 12\ \Omega $

Let’s compute R_total for problem 3:
- R1 = 2
- R2 = 4
- R3 = ?
- R4 = 1
- R5 = 1

Sum = 2+4+R3+1+1 = 8 + R3

But $ R_{\text{total}} = E / I_T = 24 / 2 = 12\ \Omega $

So:
$$
8 + R3 = 12 \Rightarrow R3 = 4\ \Omega
$$

Ah! So R3 is missing — it's blank.

Similarly, in problem 1, R3 is listed as 1 ohm — but maybe it's a typo?

Wait — in problem 1:
- R1 = 1
- R2 = 1
- R3 = 1
- R4 = 1
- Sum = 4 Ω
- E = 6 V → I = 1.5 A

But given I_T = 6 A → impossible.

Unless the resistors are not all 1 ohm.

Wait — maybe R3 is not 1 ohm?

But it says R3 = 1 ohm.

Alternatively, maybe the current is 1.5 A, not 6 A.

But it says "I_T = 6 amperes"

This suggests a typo in the worksheet.

But let’s look at problem 3, which has a blank R3 — so we can calculate it.

Let’s do problem 3 first, where we can verify.

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## ✔ Problem 3

Given:
- $ E = 24 $ V
- $ I_T = 2 $ A
- R1 = 2 Ω
- R2 = 4 Ω
- R3 = ? (blank)
- R4 = 1 Ω
- R5 = 1 Ω

Step 1: Total resistance

$$
R_{\text{total}} = \frac{E}{I_T} = \frac{24}{2} = 12\ \Omega
$$

Now sum known resistors:
$$
R1 + R2 + R4 + R5 = 2 + 4 + 1 + 1 = 8\ \Omega
$$

So:
$$
R3 = 12 - 8 = 4\ \Omega
$$

✔ So R3 = 4 Ω

Now fill in:

Currents:

In series, all currents are 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 $

- $ E1_{\text{drop}} = I_1 \times R1 = 2 \times 2 = 4 $ V
- $ E2_{\text{drop}} = 2 \times 4 = 8 $ V
- $ E5_{\text{drop}} = 2 \times 1 = 2 $ V

(We don’t need E3 or E4 drop unless asked)

Wait — the question asks for:
- $ E2_{\text{drop}} = ? $
- $ E5_{\text{drop}} = ? $

So:
- $ E2_{\text{drop}} = 8 $ V
- $ E5_{\text{drop}} = 2 $ V

Power:

$ P = I^2 \times R $

- $ P3 = I^2 \times R3 = (2)^2 \times 4 = 4 \times 4 = 16 $ W
- $ P5 = (2)^2 \times 1 = 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

✔ All consistent.

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## ✔ Problem 4

Given:
- $ E = 12 $ V
- $ I_T = 1 $ A
- R1 = 2 Ω
- R2 = 2 Ω
- R3 = 4 Ω
- R4 = 3 Ω
- R5 = 1 Ω

Step 1: Check total resistance

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

Sum of resistors:
$$
2 + 2 + 4 + 3 + 1 = 12\ \Omega ✔
$$

Perfect.

Now:

Currents:

All equal to $ I_T = 1 $ A

So:
- $ I_3 = 1 $ A
- $ I_5 = 1 $ A

Voltage Drops:

- $ E1_{\text{drop}} = I \times R1 = 1 \times 2 = 2 $ V
- $ E4_{\text{drop}} = 1 \times 3 = 3 $ V

Powers:

- $ P2 = I^2 \times R2 = 1^2 \times 2 = 2 $ W
- $ P4 = 1^2 \times 3 = 3 $ W

Total Power:

$ P_{\text{total}} = E \times I_T = 12 \times 1 = 12 $ W
Or $ I^2 \times R_{\text{total}} = 1 \times 12 = 12 $ W ✔

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

But let’s re-express them carefully.

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## ✘ Problem 1 Revisited

Given:
- $ E = 6 $ V
- $ I_T = 6 $ A → implies $ R_{\text{total}} = 1\ \Omega $
- But R1 = R2 = R3 = R4 = 1 Ω → total = 4 Ω → contradiction

So impossible.

But perhaps the current is not 6 A, but 1.5 A?

Maybe it's a typo.

Alternatively, maybe the resistors are not all 1 Ω?

Wait — maybe R3 is not 1 Ω? But it says R3 = 1 ohm.

Another possibility: the voltage is not 6 V, but 24 V?

But it says E = 6 V.

Alternatively, maybe the resistors are in parallel? But the diagram shows a single loop.

Wait — the diagram shows a series circuit.

So likely, the current is 1.5 A, not 6 A.

But it says "I_T = 6 amperes" — probably a typo.

Let’s assume that the current is actually 1.5 A, and proceed.

But the problem asks us to solve using given values.

Alternatively, maybe the resistors are not all 1 Ω?

Wait — let’s suppose R3 is 0.5 Ω or something?

But it says R3 = 1 ohm.

Wait — perhaps the 6 A is not the total current, but I1 or something?

No — it says $ I_T = 6 $ A.

So unless the battery is 24 V, not 6 V.

Wait — in problem 3, E = 24 V, I_T = 2 A → consistent.

But in problem 1, if E = 6 V, and I_T = 6 A → R_total = 1 Ω.

But we have 4 resistors of 1 Ω each → total 4 Ω.

So unless only one resistor is present, but no.

Alternatively, maybe the resistors are not all in series? But the diagram shows a single loop.

Wait — let me sketch it mentally:

- Battery E
- R1 → R2 → R3 → R4 → back to battery

Yes, series.

So total R = 4 Ω

If E = 6 V → I = 1.5 A

But given I_T = 6 A → impossible.

So must be typo.

But let’s assume that I_T = 1.5 A, and solve.

Then:

- $ I_1 = I_2 = I_3 = I_4 = 1.5 $ A

- $ E1_{\text{drop}} = 1.5 \times 1 = 1.5 $ V
- $ E2_{\text{drop}} = 1.5 \times 1 = 1.5 $ V
- $ E3_{\text{drop}} = 1.5 \times 1 = 1.5 $ V
- $ E4_{\text{drop}} = 1.5 \times 1 = 1.5 $ V

- $ P1 = (1.5)^2 \times 1 = 2.25 $ W
- $ P4 = 2.25 $ W

- $ P_{\text{total}} = 6 \times 1.5 = 9 $ W

But the problem says I_T = 6 A — so if we use that, then:

Assume I_T = 6 A, E = 6 V → R_total = 1 Ω

But we have 4 resistors → so average resistance per resistor = 0.25 Ω → but given all are 1 Ω → impossible.

So no solution possible with given values.

But wait — maybe the resistors are not all 1 Ω?

Let’s look at the original text:

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

Yes.

So likely, the current is meant to be 1.5 A, not 6 A.

Similarly, in problem 2:

- E = 3 V
- I_T = 3 A → R_total = 1 Ω
- R1 = 0.5, R2 = 1, R3 = 0.5, R4 = 1 → sum = 3 Ω → contradiction

So again, impossible.

But in problem 2:
- If E = 3 V, R_total = 3 Ω → I_T = 1 A

But given I_T = 3 A → contradiction.

So both problems 1 and 2 have inconsistent data.

But problem 3 and 4 are consistent.

Possibility: The current values are swapped?

Wait — in problem 1: I_T = 6 A, E = 6 V → needs R_total = 1 Ω

But resistors sum to 4 Ω → so maybe only one resistor is active? No.

Alternatively, maybe the resistors are not all 1 Ω?

Wait — perhaps R3 is 0.5 Ω? But it says R3 = 1 ohm.

Alternatively, maybe the voltage is 24 V in problem 1?

But it says E = 6 V.

Given the inconsistencies, perhaps the current values are typos, and should be 1.5 A and 1 A respectively.

Let’s assume that.

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## ✔ Revised Problem 1 (assuming I_T = 1.5 A)

Given:
- $ E = 6 $ V
- $ I_T = 1.5 $ A (corrected)
- R1 = R2 = R3 = R4 = 1 Ω

Then:

- $ I_1 = I_2 = I_3 = I_4 = 1.5 $ A

- $ E1_{\text{drop}} = 1.5 \times 1 = 1.5 $ V
- $ E2_{\text{drop}} = 1.5 \times 1 = 1.5 $ V
- $ E3_{\text{drop}} = 1.5 \times 1 = 1.5 $ V
- $ E4_{\text{drop}} = 1.5 \times 1 = 1.5 $ V

- $ P1 = (1.5)^2 \times 1 = 2.25 $ W
- $ P4 = 2.25 $ W

- $ P_{\text{total}} = E \times I_T = 6 \times 1.5 = 9 $ W

Or $ I^2 R_{\text{total}} = (1.5)^2 \times 4 = 2.25 \times 4 = 9 $ W ✔

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## ✔ Revised Problem 2

Given:
- $ E = 3 $ V
- $ I_T = 1 $ A (corrected)
- R1 = 0.5 Ω
- R2 = 1 Ω
- R3 = 0.5 Ω
- R4 = 1 Ω
- Total R = 0.5 + 1 + 0.5 + 1 = 3 Ω → I = 3/3 = 1 A ✔

So:

- $ I_1 = I_2 = I_3 = I_4 = 1 $ A

- $ E1_{\text{drop}} = 1 \times 0.5 = 0.5 $ V
- $ E4_{\text{drop}} = 1 \times 1 = 1 $ V

- $ P2 = I^2 R2 = 1^2 \times 1 = 1 $ W
- $ P3 = 1^2 \times 0.5 = 0.5 $ W

- $ P_{\text{total}} = E \times I_T = 3 \times 1 = 3 $ W

---

But the problem says $ I_T = 3 $ A — so still inconsistent.

But if we use $ I_T = 3 $ A, then E = 3 V → R_total = 1 Ω

But sum is 3 Ω → impossible.

So only way is to assume current is 1 A, not 3 A.

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Final Conclusion:

There are data inconsistencies in problems 1 and 2. Likely, the current values are typos.

But since problems 3 and 4 are consistent, we’ll solve them correctly.

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## ✔ Final Answers

Problem 1 (with corrected I_T = 1.5 A)

- $ I_1 = 1.5 $ A
- $ I_2 = 1.5 $ A
- $ E2_{\text{drop}} = 1.5 $ V
- $ E3_{\text{drop}} = 1.5 $ V
- $ P1 = 2.25 $ W
- $ P4 = 2.25 $ W
- $ P_{\text{total}} = 9 $ W

> Note: Given $ I_T = 6 $ A is likely a typo; should be 1.5 A.

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Problem 2 (with corrected I_T = 1 A)

- $ I_1 = 1 $ A
- $ I_3 = 1 $ A
- $ E1_{\text{drop}} = 0.5 $ V
- $ E4_{\text{drop}} = 1 $ V
- $ P2 = 1 $ W
- $ P3 = 0.5 $ W
- $ P_{\text{total}} = 3 $ W

> Note: Given $ I_T = 3 $ A is likely a typo; should be 1 A.

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Problem 3

- $ I_1 = 2 $ A
- $ I_4 = 2 $ A
- $ E2_{\text{drop}} = 8 $ V
- $ E5_{\text{drop}} = 2 $ V
- $ P3 = 16 $ W
- $ P5 = 4 $ W
- $ P_{\text{total}} = 48 $ W

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Problem 4

- $ I_3 = 1 $ A
- $ I_5 = 1 $ A
- $ E1_{\text{drop}} = 2 $ V
- $ E4_{\text{drop}} = 3 $ V
- $ P2 = 2 $ W
- $ P4 = 3 $ W
- $ P_{\text{total}} = 12 $ W

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✔ Summary Table

| Problem | I1 | I2 | I3 | I4 | I5 | E1_drop | E2_drop | E3_drop | E4_drop | E5_drop | P1 | P2 | P3 | P4 | P5 | P_total |
|--------|-----|-----|-----|-----|-----|---------|---------|---------|---------|---------|----|----|----|----|----|---------|
| 1 | 1.5 | 1.5 | 1.5 | 1.5 | — | 1.5 | 1.5 | 1.5 | 1.5 | — | 2.25 | 2.25 | 2.25 | 2.25 | — | 9 |
| 2 | 1 | 1 | 1 | 1 | — | 0.5 | — | — | 1 | — | 0.5 | 1 | 0.5 | 1 | — | 3 |
| 3 | 2 | 2 | 2 | 2 | 2 | 4 | 8 | — | — | 2 | 4 | 8 | 16 | 4 | 4 | 48 |
| 4 | 1 | 1 | 1 | 1 | 1 | 2 | — | — | 3 | — | 2 | 2 | 4 | 3 | 1 | 12 |

> Notes:
> - Problems 1 and 2 have inconsistent data; assumed current values corrected.
> - In problem 3, R3 was calculated as 4 Ω.
> - All values follow Ohm’s Law and series circuit rules.

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