Circuit analysis worksheet with problems on series and parallel resistor configurations.

Let’s solve each problem step by step. We’ll use Ohm’s Law (V = I × R) and rules for series and parallel circuits.

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Problem 1: Find I_T in the circuit

We have three resistors in parallel:
R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 6 Ω
Voltage source V_T = 6 V

In a parallel circuit, voltage across each resistor is the same as the source voltage → so each resistor has 6 V across it.

To find total current I_T, we can either:
- Find current through each resistor and add them up, OR
- Find equivalent resistance first, then use I_T = V_T / R_eq

Let’s do both to check.

Method 1: Add individual currents

I₁ = V / R₁ = 6 V / 2 Ω = 3 A
I₂ = 6 V / 3 Ω = 2 A
I₃ = 6 V / 6 Ω = 1 A

Total current I_T = I₁ + I₂ + I₃ = 3 + 2 + 1 = 6 A

Method 2: Equivalent resistance

For parallel resistors:

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃
= 1/2 + 1/3 + 1/6
= 3/6 + 2/6 + 1/6 = 6/6 = 1
→ R_eq = 1 Ω

Then I_T = V_T / R_eq = 6 V / 1 Ω = 6 A

✔ Both methods give same answer.

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Problem 2: Complete the table

Circuit: All resistors in series.
R₁ = 8 Ω, R₂ = 4 Ω, R₃ = 2 Ω, R₄ = 12 Ω
V_T = 12 V

In series:
- Current is same everywhere → I_T = I₁ = I₂ = I₃ = I₄
- Total resistance R_T = sum of all resistances
- Voltage drops add up to V_T

Step 1: Find R_T
R_T = 8 + 4 + 2 + 12 = 26 Ω

Step 2: Find I_T
I_T = V_T / R_T = 12 V / 26 Ω ≈ 0.4615 A (we’ll keep it as fraction: 12/26 = 6/13 A)

But let’s use decimal for table: ≈ 0.46 A (but better to use exact or more precise)

Actually, let’s calculate with fractions to avoid rounding errors.

I_T = 12 / 26 = 6/13 A ≈ 0.4615 A

Now find voltage drop across each:

V₁ = I × R₁ = (6/13) × 8 = 48/13 ≈ 3.692 V
V₂ = (6/13) × 4 = 24/13 ≈ 1.846 V
V₃ = (6/13) × 2 = 12/13 ≈ 0.923 V
V₄ = (6/13) × 12 = 72/13 ≈ 5.538 V

Check: Sum of voltages = 48+24+12+72 = 156 → 156/13 = 12 V ✔

So fill table:

| | R₁ | R₂ | R₃ | R₄ |
|----------|----------|----------|----------|----------|
| V_T=12V | V₁≈3.69V | V₂≈1.85V | V₃≈0.92V | V₄≈5.54V |
| R₁=8Ω | R₂=4Ω | R₃=2Ω | R₄=12Ω | |
| I_T≈0.46A| I₁≈0.46A | I₂≈0.46A | I₃≈0.46A | I₄≈0.46A |

Note: Since it's series, all currents are equal to I_T.

We can write exact values if needed, but decimals are fine for this level.

Alternatively, use fractions:

I_T = 6/13 A
V₁ = 48/13 V, V₂ = 24/13 V, V₃ = 12/13 V, V₄ = 72/13 V

But since the table doesn’t specify, we’ll use rounded decimals to two places.

Wait — let me recalculate with more precision for accuracy.

Actually, 12 ÷ 26 = 0.461538... A

So:

V₁ = 0.461538 × 8 = 3.6923 V → 3.69 V
V₂ = 0.461538 × 4 = 1.84615 V → 1.85 V
V₃ = 0.461538 × 2 = 0.923076 V → 0.92 V
V₄ = 0.461538 × 12 = 5.53846 V → 5.54 V

Sum: 3.69 + 1.85 = 5.54; +0.92 = 6.46; +5.54 = 12.00 ✔

Currents all = 0.46 A? But 0.46 × 26 = 11.96 ≈ 12 — close enough for 2 decimals.

Better to use 0.46 A for all, or be precise.

Since 12/26 = 6/13 ≈ 0.4615, we can write I = 0.46 A (rounded).

But let’s see what the problem expects. Probably acceptable to use 0.46 A.

Alternatively, leave as fraction? Unlikely for this level.

I think using 0.46 A is fine.

But wait — 0.46 × 26 = 11.96, which is off by 0.04 V. Maybe use 0.462 A?

Actually, let’s use 3 significant figures.

V_T = 12.0 V (assumed), R_T = 26.0 Ω → I_T = 12.0 / 26.0 = 0.4615 → 0.462 A

Then:

V₁ = 0.462 × 8 = 3.696 → 3.70 V
V₂ = 0.462 × 4 = 1.848 → 1.85 V
V₃ = 0.462 × 2 = 0.924 → 0.92 V
V₄ = 0.462 × 12 = 5.544 → 5.54 V

Sum: 3.70 + 1.85 = 5.55; +0.92 = 6.47; +5.54 = 12.01 — very close.

Actually, better to calculate exactly and round at end.

Use I = 12/26 = 6/13 A

V₁ = (6/13)*8 = 48/13 = 3.692307... → 3.69 V
V₂ = 24/13 = 1.84615... → 1.85 V
V₃ = 12/13 = 0.92307... → 0.92 V
V₄ = 72/13 = 5.53846... → 5.54 V

And I = 6/13 ≈ 0.46 A (if rounding to two decimals)

But 0.46 * 26 = 11.96, while actual is 12. So perhaps write I = 0.46 A, noting it’s approximate.

In many textbooks, they’d expect you to use the exact value or report consistently.

I think for this, we'll go with:

I_T = I₁ = I₂ = I₃ = I₄ = 12 V / 26 Ω = 6/13 A ≈ 0.46 A

Voltages as above.

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Problem 3: Complete the table

Circuit: Two resistors in parallel, then in series with another? Wait, looking at diagram:

It says: V_T = 10 V, and there are three resistors: R₁ = ?, R₂ = 25 Ω, R₃ = 30 Ω, and R₄ = 50 Ω? Wait, the diagram shows:

From the description: "Complete the table for this circuit" with V_T=10V, and resistors labeled R₁, R₂=25Ω, R₃=30Ω, R₄=50Ω.

Looking at the sketch: It appears that R₂ and R₃ are in parallel, and that combination is in series with R₁ and R₄? But the diagram isn't clear from text.

Wait, the user wrote: “R₁= ? , R₂=25 Ω, R₃=30 Ω, R₄=50 Ω” and V_T=10V.

And the circuit drawing: probably R₂ and R₃ are in parallel, and that parallel combo is in series with R₁ and R₄? But that would be four resistors.

Perhaps it's R₁ in series with a parallel combination of R₂ and R₃, and R₄ is not there? But it lists R₄=50Ω.

Looking back at original image description: In problem 3, it says “R₁= ? , R₂=25 Ω, R₃=30 Ω, R₄=50 Ω” and V_T=10V.

And the circuit: likely R₂ and R₃ are in parallel, and that parallel group is in series with R₁ and R₄? But that seems odd.

Perhaps it's a typo or mislabeling. Another possibility: maybe R₄ is the equivalent or something.

Wait, in the table, it has columns for R₁, R₂, R₃, R₄ — so four resistors.

But in the circuit description, it might be that R₂ and R₃ are in parallel, and R₁ and R₄ are in series with that combination.

Assume the circuit is: Battery — R₁ — [parallel: R₂ and R₃] — R₄ — back to battery.

That makes sense.

So total resistance R_T = R₁ + R_parallel + R₄

Where R_parallel = (R₂ * R₃)/(R₂ + R₃) = (25 * 30)/(25+30) = 750 / 55 ≈ 13.636 Ω

But we don't know R₁ yet. The table asks for R₁, and also voltages and currents.

We have V_T = 10 V.

But we need more information. The table has blanks for V₁, V₂, V₃, V₄, I₁, I₂, I₃, I, and R₁.

This is underdetermined unless we assume something.

Perhaps R₁ is to be found, but how?

Another thought: maybe the circuit is only three resistors, and R₄ is a mistake? But it's listed.

Looking back at the user's input: "R₁= ? , R₂=25 Ω, R₃=30 Ω, R₄=50 Ω"

And in the circuit diagram description, it might be that R₂ and R₃ are in parallel, and R₁ is in series with them, and R₄ is not present? But why list it?

Perhaps R₄ is the total or something. I think there might be a confusion.

Another idea: perhaps the circuit is R₁ in series with a parallel combination of R₂, R₃, and R₄? But that would be three in parallel.

The user said: "R₂=25 Ω, R₃=30 Ω, R₄=50 Ω" — so three resistors in parallel? But then R₁ is separate.

Let me read the original problem again from the image description.

In the initial prompt, for problem 3: "Complete the table for this circuit" with V_T=10V, and resistors R₁=?, R₂=25Ω, R₃=30Ω, R₄=50Ω.

And the circuit drawing: from the ASCII art, it shows a battery, then a resistor R₁, then a parallel branch with two resistors (probably R₂ and R₃), and then another resistor R₄? But that would be four.

Perhaps it's R₁ in series with the parallel combination of R₂ and R₃, and R₄ is not used? But it's listed.

Maybe R₄ is the equivalent resistance or something. I think there's a mistake in interpretation.

Another possibility: in some diagrams, R₄ might be the load or something, but here it's listed as a resistor.

Perhaps the circuit is: battery — R₁ — then parallel: R₂ and R₃ — and that's it, and R₄ is a typo. But the table has R₄.

Let's look at the table structure:

| | R₁ | R₂ | R₃ | R₄ |
|----------|----------|----------|----------|----------|
| V_T=10V | V₁= | V₂= | V₃= | V₄= |
| R₁= | R₂=25Ω | R₃=30Ω | R₄=50Ω | |
| I_T= | I₁= | I₂= | I₃= | I₄= |

So four resistors. Likely, the circuit is R₁ in series with a parallel combination of R₂, R₃, and R₄? But that would be three in parallel.

Or perhaps R₂ and R₃ are in parallel, and that is in series with R₁ and R₄.

I think that's the most reasonable assumption.

So let's assume: R_total = R₁ + R_parallel + R₄, where R_parallel = (R₂ * R₃)/(R₂ + R₃) = (25*30)/(25+30) = 750/55 = 150/11 ≈ 13.636 Ω

But we don't know R₁, and we have V_T=10V, but no other given values. How can we find anything?

Unless... perhaps the current or voltage is given implicitly. Or maybe R₁ is to be found such that something is true, but nothing is specified.

This is problematic. Perhaps in the diagram, it's shown that the voltage across the parallel part is given or something, but not stated.

Another thought: maybe "R₄=50Ω" is a mistake, and it's supposed to be the total or something. Or perhaps R₄ is not a resistor but the equivalent.

I recall that in some problems, they label the equivalent as R₄, but here it's listed with others.

Perhaps the circuit is only three resistors: R₁ in series with parallel R₂ and R₃, and R₄ is not there, but the table has it by mistake.

But the user specifically wrote "R₄=50Ω", so it must be included.

Let's try to search for standard problems. Perhaps R₂ and R₃ are in parallel, and R₁ and R₄ are in series with that, but we need another condition.

Maybe the total current is given, but it's not.

Another idea: perhaps "I_T=" is to be filled, but we have no value.

This is confusing. Let's look back at the original image description.

In the user's message, for problem 3, it says: "Complete the table for this circuit" and then "V_T=10 V", "R₁= ?", "R₂=25 Ω", "R₃=30 Ω", "R₄=50 Ω".

And the circuit drawing: from the ASCII, it might be that R₂ and R₃ are in parallel, and R₁ is in series with them, and R₄ is the fourth resistor in series or something.

Perhaps it's a series circuit with four resistors, but then why mention parallel in the diagram? The user said "for this circuit" and described it as having a parallel section.

In the initial problem statement, for problem 3, it says: "Complete the table for this circuit" and the circuit has a parallel combination.

From the way it's written: "R₁= ? , R₂=25 Ω, R₃=30 Ω, R₄=50 Ω" and V_T=10V.

Perhaps R₄ is the equivalent resistance of the parallel part? But that doesn't make sense because R₂ and R₃ are given, and their parallel equivalent is not 50Ω.

(25*30)/(25+30) = 750/55 ≈ 13.6, not 50.

Another possibility: maybe R₂, R₃, R₄ are in parallel, and R₁ is in series with them.

Let me try that.

Assume: R_parallel = 1/(1/25 + 1/30 + 1/50)

Calculate: 1/25 = 0.04, 1/30 ≈ 0.0333, 1/50 = 0.02, sum = 0.04+0.0333+0.02 = 0.0933, so R_parallel = 1/0.0933 ≈ 10.714 Ω

Then R_T = R₁ + 10.714

But still unknown R₁.

Perhaps the voltage across the parallel part is given, but it's not.

I think there might be a missing piece. Perhaps in the diagram, it's shown that the current or voltage is known for one part.

Maybe "R₄=50Ω" is a red herring, or perhaps it's the total resistance.

Another idea: perhaps R₄ is not a resistor, but the label for the equivalent, but the table has it as a column.

Let's look at the table: it has R₁, R₂, R₃, R₄ as separate columns, so likely four distinct resistors.

Perhaps the circuit is R₁ in series with R₄, and in between, there is a parallel combination of R₂ and R₃.

That is common.

So let's assume that.

So the circuit is: battery — R₁ — node A — then R₂ and R₃ in parallel — node B — R₄ — back to battery.

So total resistance R_T = R₁ + R_parallel + R₄, with R_parallel = (R₂*R₃)/(R₂+R₃) = (25*30)/(55) = 750/55 = 150/11 ≈ 13.636 Ω

R₄ = 50 Ω

R₁ = ?

V_T = 10 V

But we have two unknowns: R₁ and the currents/voltages.

Unless... perhaps the table is to be filled with expressions, but that seems unlikely.

Maybe in the diagram, it's indicated that the voltage across R₂ or something is given, but not stated.

Perhaps "I_T=" is to be calculated, but we need R₁.

I think there might be a mistake in the problem or my understanding.

Another thought: perhaps "R₄=50Ω" is the value of R₁, but it's listed as R₄.

Or perhaps R₁ is 50Ω, but it's written as R₄.

Let's check the user's input: "R₁= ? , R₂=25 Ω, R₃=30 Ω, R₄=50 Ω"

So R₁ is unknown, R₄ is 50Ω.

Perhaps the total resistance is given, but it's not.

Maybe the current is the same, but in series-parallel, it's not.

Let's assume that the parallel combination of R₂ and R₃ has a certain voltage, but not given.

Perhaps from the context, we can find that the voltage across the parallel part is proportional, but without additional info, it's impossible.

Unless... in some problems, they imply that the resistors are connected in a specific way, and perhaps R₁ is to be found such that the total current is integer or something, but not specified.

I recall that in the initial problem, for problem 3, it might be that R₂ and R₃ are in parallel, and that combination is in series with R₁, and R₄ is not present, but the table has R₄ by mistake.

Perhaps "R₄=50Ω" is the equivalent resistance of the parallel part, but (25*30)/(25+30) = 750/55 ≈ 13.6, not 50.

Another idea: perhaps R₄ is in series with the parallel combination, and R₁ is not there, but it's listed.

Let's calculate the parallel equivalent of R₂ and R₃: 1/R_p = 1/25 + 1/30 = (6+5)/150 = 11/150, so R_p = 150/11 ≈ 13.636 Ω

If R₄ = 50 Ω is in series, then R_T = R₁ + 13.636 + 50 = R₁ + 63.636

V_T = 10 V, so I_T = 10 / (R₁ + 63.636)

But still unknown.

Perhaps R₁ is 0, but that doesn't make sense.

Maybe the circuit is only the parallel combination of R₂, R₃, R, and R₁ is the equivalent or something.

Let's try that. Suppose R₂, R₃, R₄ are in parallel, and R₁ is the total resistance or something.

But the table has R₁ as a resistor.

Perhaps "R₁= ?" means we need to find the equivalent resistance, but it's labeled as R₁.

I think there's a confusion in labeling.

Let me look for similar problems online or standard setups.

Perhaps in the diagram, it's shown that the voltage across R₁ is given, but not stated.

Another possibility: perhaps "V_T=10V" is across the entire circuit, and we need to find R₁ such that the current is nice, but not specified.

Maybe the table is to be filled with formulas, but that seems advanced for this level.

Let's read the user's message carefully: "Complete the table for this circuit" and then the values.

Perhaps for problem 3, the circuit is R₁ in series with a parallel combination of R₂ and R₃, and R₄ is not used, but it's listed by mistake. Or perhaps R₄ is the parallel equivalent.

But (25*30)/(25+30) = 750/55 = 150/11 ≈ 13.6, not 50.

Unless R₄ is 50Ω and it's in series, and R₁ is to be found from other conditions.

I think I need to make an assumption.

Let me assume that the circuit is: battery — R₁ — then parallel: R₂ and R₃ — and that's it, and R₄ is a typo or not used. But the table has R₄, so perhaps omit it or set to 0.

But that's not good.

Another idea: perhaps "R₄=50Ω" is the value of the parallel equivalent, but it's given as 50, while calculation gives 13.6, so not.

Unless R₂ and R₃ are not the only ones.

Let's calculate if R₂, R₃, R₄ are in parallel: 1/R_p = 1/25 + 1/30 + 1/50 = (6 + 5 + 3)/150 = 14/150 = 7/75, so R_p = 75/7 ≈ 10.714 Ω

Then if R₁ is in series, R_T = R₁ + 10.714

V_T = 10 V, so I_T = 10 / (R₁ + 10.714)

Still unknown.

Perhaps R₁ is given as 0, but not.

I recall that in some problems, they have the total resistance given, but here it's not.

Perhaps from the context of the worksheet, but we don't have it.

Another thought: in the table, for R₁, it's "?", and for others, values are given, so likely R₁ is to be calculated from the circuit laws, but we need more data.

Unless the current is the same, but in parallel, it's not.

Perhaps the voltage across R₂ is given, but not.

I think there might be a mistake in the problem transcription.

Let's look back at the user's initial message: "3. Complete the table for this circuit" and then "V_T=10 V", "R₁= ?", "R₂=25 Ω", "R₃=30 Ω", "R₄=50 Ω".

And in the circuit description, it might be that R₂ and R₃ are in parallel, and R₁ and R₄ are in series with that, but perhaps R₁ is 0 or something.

Perhaps "R₄=50Ω" is the total resistance, but then R_T = 50 Ω, V_T = 10 V, so I_T = 10/50 = 0.2 A.

Then if R₂ and R₃ are in parallel, R_p = (25*30)/(25+30) = 750/55 = 150/11 ≈ 13.636 Ω

Then if R_T = R₁ + R_p + R₄, but R₄ is already included? Confusing.

Assume that R_T = 50 Ω, and it consists of R₁ in series with R_p, where R_p is parallel of R₂ and R₃.

So R_T = R₁ + R_p = R₁ + 150/11 = 50

So R₁ = 50 - 150/11 = (550 - 150)/11 = 400/11 ≈ 36.36 Ω

Then I_T = V_T / R_T = 10 / 50 = 0.2 A

Then voltage across R₁: V₁ = I_T * R₁ = 0.2 * 400/11 = 80/11 ≈ 7.27 V

Voltage across parallel part: V_p = I_T * R_p = 0.2 * 150/11 = 30/11 ≈ 2.73 V

Since R₂ and R₃ are in parallel, V₂ = V₃ = V_p = 2.73 V

Then I₂ = V₂ / R₂ = 2.73 / 25 = 0.1092 A

I₃ = 2.73 / 30 = 0.091 A

Sum I₂ + I₃ = 0.1092 + 0.091 = 0.2002 A ≈ I_T, good.

But what about R₄? In this assumption, R₄ is not used, or perhaps R₄ is the parallel equivalent, but it's given as 50Ω, which is R_T.

In the table, R₄ is listed as 50Ω, so perhaps R₄ is the total resistance, but typically R_T is not labeled as R₄.

Perhaps in the circuit, R₄ is not a physical resistor, but the equivalent.

But the table has it as a column, so likely it's a resistor.

Perhaps for R₄, since it's not in the circuit, we leave blank, but that's not satisfactory.

Another possibility: perhaps the circuit is R₁ in series with R₄, and the parallel combination of R₂ and R₃ is across R₄ or something, but complicated.

I think for the sake of progressing, I'll assume that R₄ is the total resistance or something, but let's see the answer format.

Perhaps in problem 3, "R₄=50Ω" is a mistake, and it's supposed to be the value of R₁ or something.

Let's calculate if R₁ is 50Ω, then with R_p = 13.636Ω, R_T = 50 + 13.636 = 63.636Ω, I_T = 10/63.636 ≈ 0.157 A, etc, but then R₄ is not used.

I think the most reasonable assumption is that the circuit has R₁ in series with the parallel combination of R₂ and R₃, and R₄ is not present, but since it's in the table, perhaps set R₄=0 or omit.

But to match the table, perhaps R₄ is the parallel equivalent, but it's given as 50, while it should be 13.6.

Unless the values are different.

Another idea: perhaps "R₄=50Ω" is the value for the series resistor, and R₁ is the parallel equivalent or something.

Let's swap: suppose R₁ is the parallel equivalent of R₂ and R₃, but R₂ and R₃ are given, so R₁ = (25*30)/(25+30) = 750/55 = 150/11 ≈ 13.636 Ω, but the table has R₁= ? , and R₄=50Ω, so perhaps R₄ is in series.

Then R_T = R₁ + R₄ = 13.636 + 50 = 63.636 Ω

I_T = 10 / 63.636 ≈ 0.157 A

V₁ = I_T * R₁ = 0.157 * 13.636 ≈ 2.14 V (voltage across parallel part)

V₄ = I_T * R₄ = 0.157 * 50 = 7.85 V

Then for R₂ and R₃, since in parallel, V₂ = V₃ = V₁ = 2.14 V

I₂ = 2.14 / 25 = 0.0856 A

I₃ = 2.14 / 30 = 0.0713 A

Sum 0.0856 + 0.0713 = 0.1569 A ≈ I_T, good.

Then in the table, R₁ = 13.636 Ω, but the table has R₁= ? , and R₄=50Ω, so perhaps R₁ is to be calculated as the parallel equivalent.

But typically, R₁ is a separate resistor.

Perhaps in the diagram, R₁ is the parallel combination, but usually it's labeled as such.

I think for the sake of completing, I'll assume that the circuit is: R₄ in series with the parallel combination of R₂ and R₃, and R₁ is the equivalent resistance of the parallel part or something, but the table has R₁ as a resistor.

Perhaps "R₁= ?" means we need to find the resistance of the first resistor, but in this case, if R₄ is 50Ω, and it's in series, then R₁ might be the parallel part.

Let's define: let R_parallel = R₁ = (R₂*R₃)/(R₂+R₃) = (25*30)/(55) = 750/55 = 150/11 ≈ 13.636 Ω

Then R_T = R₁ + R₄ = 13.636 + 50 = 63.636 Ω

I_T = V_T / R_T = 10 / 63.636 = 10 / (700/11) wait, 63.636 = 700/11? 700/11 = 63.636 yes, since 11*63.636=700.

R_T = R₁ + R₄ = 150/11 + 50 = 150/11 + 550/11 = 700/11 Ω

I_T = 10 / (700/11) = 10 * 11 / 700 = 110/700 = 11/70 A ≈ 0.1571 A

V₁ = I_T * R₁ = (11/70) * (150/11) = 150/70 = 15/7 ≈ 2.1429 V (this is the voltage across the parallel combination)

V₄ = I_T * R₄ = (11/70) * 50 = 550/70 = 55/7 ≈ 7.8571 V

For R₂ and R₃, since in parallel, V₂ = V₃ = V₁ = 15/7 V ≈ 2.1429 V

I₂ = V₂ / R₂ = (15/7) / 25 = 15/(7*25) = 15/175 = 3/35 A ≈ 0.0857 A

I₃ = (15/7) / 30 = 15/(7*30) = 15/210 = 1/14 A ≈ 0.0714 A

Sum I₂ + I₃ = 3/35 + 1/14 = 6/70 + 5/70 = 11/70 A = I_T, good.

Now, in the table, R₁ is the resistance of the parallel combination, so R₁ = 150/11 Ω ≈ 13.64 Ω

R₂ = 25 Ω, R₃ = 30 Ω, R₄ = 50 Ω

V₁ = 15/7 V ≈ 2.14 V (voltage across R₁, which is the parallel combo)

V₂ = 15/7 V ≈ 2.14 V

V₃ = 15/7 V ≈ 2.14 V

V₄ = 55/7 V ≈ 7.86 V

I_T = 11/70 A ≈ 0.157 A

I₁ = I_T = 11/70 A (since R₁ is in series with R₄, and the parallel combo is treated as R₁, so current through R₁ is I_T)

In this interpretation, R₁ is not a physical resistor but the equivalent resistance of the parallel branch. But in the table, it's listed as a resistor, so perhaps it's acceptable.

I₁ is the current through the equivalent resistor R₁, which is the same as I_T, since it's in series.

I₂ and I₃ are the currents through R₂ and R₃.

I₄ = I_T, since R₄ is in series.

So let's fill the table with this.

R₁ = 150/11 Ω ≈ 13.64 Ω

V₁ = 15/7 V ≈ 2.14 V

I₁ = 11/70 A ≈ 0.157 A

R₂ = 25 Ω

V₂ = 15/7 V ≈ 2.14 V

I₂ = 3/35 A ≈ 0.0857 A

R₃ = 30 Ω

V₃ = 15/7 V ≈ 2.14 V

I₃ = 1/14 A ≈ 0.0714 A

R₄ = 50 Ω

V₄ = 55/7 V ≈ 7.86 V

I₄ = 11/70 A ≈ 0.157 A

I_T = 11/70 A ≈ 0.157 A

This seems consistent.

Perhaps this is what is intended.

So for problem 3, we'll go with this.

Now to summarize all answers.

Final Answer:

Problem 1:
I_T = 6 A

Problem 2:
R_T = 26 Ω
I_T = I₁ = I₂ = I₃ = I₄ = 12/26 = 6/13 A ≈ 0.46 A
V₁ = (6/13)*8 = 48/13 V ≈ 3.69 V
V₂ = (6/13)*4 = 24/13 V ≈ 1.85 V
V₃ = (6/13)*2 = 12/13 V ≈ 0.92 V
V₄ = (6/13)*12 = 72/13 V ≈ 5.54 V

Problem 3:
R₁ = (25*30)/(25+30) = 750/55 = 150/11 Ω ≈ 13.64 Ω
I_T = I₁ = I₄ = 10 / (R₁ + R₄) = 10 / (150/11 + 50) = 10 / (150/11 + 550/11) = 10 / (700/11) = 110/700 = 11/70 A ≈ 0.157 A
V₁ = I_T * R₁ = (11/70) * (150/11) = 150/70 = 15/7 V ≈ 2.14 V
V₂ = V₁ = 15/7 V ≈ 2.14 V (since parallel)
V₃ = V₁ = 15/7 V ≈ 2.14 V
V₄ = I_T * R₄ = (11/70) * 50 = 550/70 = 55/7 V ≈ 7.86 V
I₂ = V₂ / R₂ = (15/7) / 25 = 15/(175) = 3/35 A ≈ 0.0857 A
I₃ = V₃ / R₃ = (15/7) / 30 = 15/210 = 1/14 A ≈ 0.0714 A

Now for the final answer box, since it's multiple parts, we'll list them.

But the instruction is to provide the final correct answer after all explanations, so perhaps box the key results.

Since the user may expect the answers for each problem, we'll state them clearly.

For Problem 1: I_T = 6 A

For Problem 2:
- I_T = I₁ = I₂ = I₃ = I₄ = \frac{6}{13} A \approx 0.46 A
- V₁ = \frac{48}{13} V \approx 3.69 V
- V₂ = \frac{24}{13} V \approx 1.85 V
- V₃ = \frac{12}{13} V \approx 0.92 V
- V₄ = \frac{72}{13} V \approx 5.54 V

For Problem 3:
- R₁ = \frac{150}{11} \Omega \approx 13.64 \Omega
- I_T = I₁ = I₄ = \frac{11}{70} A \approx 0.157 A
- V₁ = V₂ = V₃ = \frac{15}{7} V \approx 2.14 V
- V₄ = \frac{55}{7} V \approx 7.86 V
- I₂ = \frac{3}{35} A \approx 0.0857 A
- I₃ = \frac{1}{14} A \approx 0.0714 A

To make it neat, perhaps use decimals rounded to two places for simplicity, as often done in such worksheets.

For Problem 2:
I_T = 0.46 A
V₁ = 3.69 V
V₂ = 1.85 V
V₃ = 0.92 V
V₄ = 5.54 V

For Problem 3:
R₁ = 13.64 Ω
I_T = 0.16 A (rounded)
But 11/70 = 0.15714, so 0.16 A if rounded to two decimals, but let's use three for accuracy.

Since the voltages are given as 10V, probably expect two decimals.

In Problem 2, V_T=12V, so perhaps two decimals.

For Problem 3, V_T=10V, so two decimals.

So for Problem 3:
R₁ = 13.64 Ω
I_T = 0.16 A (but 0.157 is closer to 0.16, but let's calculate: 11/70 = 0.157142..., so 0.16 A if rounded, but perhaps keep as 0.157 or use fraction.

To be precise, we can use the exact values or rounded.

I think for school level, rounded to two decimals is fine.

So for Problem 3:
R₁ = 13.64 Ω
I_T = 0.16 A (since 0.157 rounds to 0.16)
V₁ = 2.14 V
V₂ = 2.14 V
V₃ = 2.14 V
V₄ = 7.86 V
I₂ = 0.09 A (0.0857 rounds to 0.09)
I₃ = 0.07 A (0.0714 rounds to 0.07)
I₁ = 0.16 A
I₄ = 0.16 A

But sum I₂ + I₃ = 0.09 + 0.07 = 0.16 A, good.

V₁ + V₄ = 2.14 + 7.86 = 10.00 V, good.

So acceptable.

Now for the final answer.

Since the user may want the answers boxed, and there are three problems, we'll provide them separately.

But the instruction is to have a single Final Answer section.

Perhaps list the key results.

Given that, I'll write the final answers as:

For problem 1: \( I_T = 6 \text{A} \)

For problem 2:
- \( I_T = I_1 = I_2 = I_3 = I_4 = 0.46 \text{A} \)
- \( V_1 = 3.69 \text{V}, V_2 = 1.85 \text{V}, V_3 = 0.92 \text{V}, V_4 = 5.54 \text{V} \)

For problem 3:
- \( R_1 = 13.64 \Omega \)
- \( I_T = I_1 = I_4 = 0.16 \text{A} \)
- \( V_1 = V_2 = V_3 = 2.14 \text{V} \)
- \( V_4 = 7.86 \text{V} \)
- \( I_2 = 0.09 \text{A}, I_3 = 0.07 \text{A} \)

To save space, perhaps just state the values.

Since the table needs to be completed, but for final answer, the numerical values.

I think this is as good as it gets.

So Final Answer:

Problem 1: \( I_T = 6 \text{A} \)

Problem 2:
- Currents: \( I_T = I_1 = I_2 = I_3 = I_4 = 0.46 \text{A} \)
- Voltages: \( V_1 = 3.69 \text{V}, V_2 = 1.85 \text{V}, V_3 = 0.92 \text{V}, V_4 = 5.54 \text{V} \)

Problem 3:
- \( R_1 = 13.64 \Omega \)
- \( I_T = I_1 = I_4 = 0.16 \text{A} \)
- \( V_1 = V_2 = V_3 = 2.14 \text{V} \)
- \( V_4 = 7.86 \text{V} \)
- \( I_2 = 0.09 \text{A}, I_3 = 0.07 \text{A} \)

Note: For Problem 3, the assumption is that R₁ represents the equivalent resistance of the parallel combination of R₂ and R₃, and R₄ is in series with it. This is based on the given values and typical circuit configurations.

Now for the box.

Since it's long, perhaps just state the main answers.

I think for the purpose, we'll box the final numerical answers as per the problems.

So:

Final Answer:
\[ \boxed{\begin{array}{c} \text{Problem 1: } I_T = 6 \text{A} \\ \\ \text{Problem 2:} \\ I_T = I_1 = I_2 = I_3 = I_4 = 0.46 \text{A} \\ V_1 = 3.69 \text{V}, V_2 = 1.85 \text{V}, V_3 = 0.92 \text{V}, V_4 = 5.54 \text{V} \\ \\ \text{Problem 3:} \\ R_1 = 13.64 \Omega \\ I_T = I_1 = I_4 = 0.16 \text{A} \\ V_1 = V_2 = V_3 = 2.14 \text{V} \\ V_4 = 7.86 \text{V} \\ I_2 = 0.09 \text{A}, I_3 = 0.07 \text{A} \end{array}} \]

To make it cleaner, perhaps use separate boxes, but the instruction is one Final Answer section.

I think this is acceptable.

Note: In Problem 3, the values are rounded to two decimal places for practicality. Exact values are fractions as calculated earlier.