Parallel Circuit Worksheet #3: Six circuit diagrams with specified voltages, currents, and resistances for analyzing series-parallel circuits.

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

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Problem 1:

Circuit: E → splits into two branches:
- Branch 1: R1 in series with (R2 || R3)
- Branch 2: R4 alone

Given:
E = 12 V
It = 2 A
R1 = 6 Ω
R2 = 12 Ω
R3 = ?
R4 = 12 Ω

First, find total resistance:
Rt = E / It = 12 V / 2 A = 6 Ω

Now, let’s break down the circuit.

Branch 2 (R4) is directly across the source → voltage across R4 = 12 V
Current through R4: I4 = V/R4 = 12/12 = 1 A

Total current = 2 A → so current through Branch 1 = 2 - 1 = 1 A

Branch 1 has R1 (6 Ω) in series with parallel combo of R2 and R3.

Voltage drop across R1: V1 = I × R1 = 1 A × 6 Ω = 6 V
So voltage across (R2 || R3) = 12 - 6 = 6 V

Since R2 = 12 Ω, current through R2: I2 = 6 V / 12 Ω = 0.5 A
Current through R3: I3 = total branch current - I2 = 1 - 0.5 = 0.5 A
So R3 = V / I3 = 6 V / 0.5 A = 12 Ω

✔ R3 = 12 ohms

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Problem 2:

Circuit: E → R1 → then splits into two branches:
- Branch A: R2 alone
- Branch B: R3 and R4 in series

Given:
E = 24 V
It = ?
R1 = 2 Ω
R2 = 8 Ω
R3 = 4 Ω
R4 = 4 Ω

First, find resistance of Branch B: R3 + R4 = 4 + 4 = 8 Ω

Now, Branch A (R2=8Ω) and Branch B (8Ω) are in parallel → equivalent resistance = (8×8)/(8+8) = 64/16 = 4 Ω

Add R1 in series: Rt = R1 + 4 = 2 + 4 = 6 Ω

Total current: It = E / Rt = 24 V / 6 Ω = 4 amperes

✔ It = 4 amperes

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Problem 3:

Circuit: E → R3 → then splits into two branches:
- Branch A: R1 alone
- Branch B: R2 alone
Then both recombine → go through R4 → back to E

Wait — looking at diagram: Actually, it's E → R3 → then splits to R1 and R2 (parallel), then they join → go through R4 → back to E.

So: R1 || R2, then in series with R3 and R4.

Given:
It = 1 A
R1 = ?
R2 = 12 Ω
R3 = 12 Ω
R4 = 4 Ω

Total resistance: Rt = R3 + (R1 || R2) + R4

But we know It = 1 A, and E is unknown? Wait — actually, E is blank, but we can find R1 using the fact that total current is 1 A and we have other values.

Actually, since It = 1 A flows through R3 and R4, and also through the parallel combo.

Voltage drop across R3: V3 = I × R3 = 1 × 12 = 12 V
Voltage drop across R4: V4 = 1 × 4 = 4 V
So voltage across parallel combo (R1 || R2) = E - 12 - 4 = E - 16

But we don’t know E yet.

Wait — perhaps we can find equivalent resistance first.

Total resistance Rt = R3 + R_parallel + R4 = 12 + R_parallel + 4 = 16 + R_parallel

But It = 1 A → E = It × Rt = 1 × (16 + R_parallel) = 16 + R_parallel

Also, voltage across parallel combo = I_parallel × R_parallel = 1 × R_parallel (since same current enters parallel combo)

But also, voltage across R2 = I2 × R2, and I2 = V_parallel / R2 = R_parallel / 12

Similarly, I1 = R_parallel / R1

And I1 + I2 = 1 A → R_parallel / R1 + R_parallel / 12 = 1

Factor R_parallel: R_parallel (1/R1 + 1/12) = 1

But R_parallel = (R1 × 12) / (R1 + 12)

Substitute:

[(R1 × 12)/(R1 + 12)] × [1/R1 + 1/12] = 1

Simplify inside: 1/R1 + 1/12 = (12 + R1)/(12 R1)

So:

[12 R1 / (R1 + 12)] × [(12 + R1)/(12 R1)] = 1

Notice: (R1 + 12) cancels, 12 R1 cancels → left with 1 = 1

That means our equation is always true? That can’t be.

Wait — I think I made a mistake. The total current is 1 A, which goes through R3 and R4, and also through the parallel combination. So yes, current through parallel combo is 1 A.

Voltage across parallel combo = I × R_parallel = 1 × R_parallel

But also, since R1 and R2 are in parallel, voltage across them is same.

Let Vp = voltage across parallel combo.

Then I1 = Vp / R1, I2 = Vp / 12

I1 + I2 = 1 → Vp (1/R1 + 1/12) = 1

Also, total voltage E = V_R3 + Vp + V_R4 = 1×12 + Vp + 1×4 = 16 + Vp

But we don't know E or Vp.

Wait — perhaps we can express everything in terms of R1.

From above: Vp (1/R1 + 1/12) = 1 → Vp = 1 / (1/R1 + 1/12) = 1 / [(12 + R1)/(12 R1)] = 12 R1 / (12 + R1)

Now, total resistance Rt = R3 + R_parallel + R4 = 12 + [R1*12/(R1+12)] + 4 = 16 + 12 R1/(R1+12)

But It = 1 A, so E = It * Rt = 16 + 12 R1/(R1+12)

But also E = 16 + Vp = 16 + 12 R1/(R1+12) — same thing.

So we need another way.

Perhaps the problem expects us to realize that with It=1A, and given R2,R3,R4, we can find R1 if we assume something? But no.

Wait — let's look back at the diagram description. In problem 3, it says:

"E = ______
It = 1 ampere
R1 = ______
R2 = 12 ohms
R3 = 12 ohms
R4 = 4 ohms"

So both E and R1 are unknown. But we have only one equation. That suggests maybe I misread the circuit.

Looking again at the user's image description for problem 3:

"3. [circuit] E → R3 → then splits to R1 and R2 (parallel) → then joins → R4 → back to E"

Yes.

But with It=1A, and R2,R3,R4 known, R1 is still free? That doesn't make sense for a homework problem.

Unless... perhaps the "It" is not the total current? But it says "It = 1 ampere", and in context, It usually means total current.

Maybe I need to use the fact that in parallel, voltages are equal, but I think I did that.

Another approach: let's denote R_parallel = Rp = (R1 * 12) / (R1 + 12)

Total resistance Rt = 12 + Rp + 4 = 16 + Rp

E = It * Rt = 1 * (16 + Rp) = 16 + Rp

Voltage across Rp is Vp = It * Rp = Rp (since It=1A through it)

But also, Vp = I2 * R2 = I2 * 12

And I2 = Vp / 12 = Rp / 12

Similarly, I1 = Vp / R1 = Rp / R1

I1 + I2 = 1 → Rp / R1 + Rp / 12 = 1

As before.

Rp (1/R1 + 1/12) = 1

But Rp = 12 R1 / (R1 + 12)

So [12 R1 / (R1 + 12)] * [ (12 + R1)/(12 R1) ] = 1 → 1 = 1

This is an identity, meaning that for any R1, as long as E is adjusted accordingly, It=1A is satisfied. But that can't be for a specific answer.

Perhaps I misinterpreted the circuit. Let me read the user's text again.

For problem 3: "E = ______ It = 1 ampere R1 = ______ R2 = 12 ohms R3 = 12 ohms R4 = 4 ohms"

And the circuit is described as having R3, then R1 and R2 in parallel, then R4.

But maybe R4 is not in series? Or perhaps the diagram shows something else.

Another possibility: perhaps "It" is the current through R1 or something, but it says "It", which typically means total current.

Maybe in this context, "It" is given, and we need to find R1 and E, but there are two unknowns.

Unless... let's calculate the voltage drops.

Suppose we let R1 be x.

Then Rp = (x * 12) / (x + 12)

Rt = 12 + 4 + Rp = 16 + Rp

E = 1 * Rt = 16 + Rp

Voltage across Rp = 1 * Rp = Rp

Current through R2 = Rp / 12

Current through R1 = Rp / x

Sum = Rp/12 + Rp/x = 1

So Rp (1/12 + 1/x) = 1

But Rp = 12x/(x+12)

So [12x/(x+12)] * [(x + 12)/(12x)] = 1 → 1=1

Indeed, it's always true. So for any R1, if E is set to 16 + Rp, then It=1A.

But that means R1 can be anything, which is not possible for a homework problem.

Perhaps I have the circuit wrong. Let me look at the user's initial description.

In the user's message, for problem 3, it says:

"3. [circuit] E → R3 → then splits to R1 and R2 (parallel) → then joins → R4 → back to E"

But in the text, it might be different. Perhaps R4 is in parallel or something.

Another thought: maybe "It" is not the total current, but the current through R1 or R2. But it says "It", and in problem 1,2,4,5,6, "It" is used for total current.

In problem 4, It is given, and we calculate.

Perhaps for problem 3, we need to assume that the voltage is such that, but no.

Let's try to plug in a value. Suppose R1 = 12 ohms.

Then Rp = (12*12)/(12+12) = 144/24 = 6 ohms

Rt = 12 + 6 + 4 = 22 ohms

E = 1 * 22 = 22 V

Voltage across Rp = 1 * 6 = 6 V

Current through R2 = 6/12 = 0.5 A

Current through R1 = 6/12 = 0.5 A

Sum = 1 A, good.

If R1 = 6 ohms, Rp = (6*12)/(6+12) = 72/18 = 4 ohms

Rt = 12 + 4 + 4 = 20 ohms

E = 20 V

Vp = 4 V

I2 = 4/12 = 1/3 A

I1 = 4/6 = 2/3 A

Sum = 1 A, good.

So R1 can be any value, as long as E is adjusted. But that can't be.

Unless the problem has a typo, or I misread the circuit.

Let's look at the user's text for problem 3: "E = ______ It = 1 ampere R1 = ______ R2 = 12 ohms R3 = 12 ohms R4 = 4 ohms"

Perhaps "It" is the current through R1? But it's labeled "It", which is confusing.

Maybe in the diagram, R4 is not in series. Let me think differently.

Another possibility: perhaps the circuit is E → R3 → then R1 and R2 in parallel, and R4 is in series with the whole thing, but maybe R4 is after the parallel, which I have.

Perhaps "It" is the current through R3 or R4, but since it's series, it's the same as total current.

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

Let's skip to problem 4 and come back.

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Problem 4:

Circuit: E → R1 → then splits into two branches:
- Branch A: R2 in series with R4
- Branch B: R3 in series with R5? Wait, let's see.

From user's description: "4. [circuit] E → R1 → then splits to: one branch has R2 and R4 in series, other branch has R3 and R5 in series? But it says R3 and R5, but in the text: "R2 = 12 ohms R3 = 12 ohms R5 = 12 ohms R1 = 6 ohms R4 = 6 ohms"

And "E = 24 volts It = ?"

So likely: after R1, it splits to two branches:
- Branch 1: R2 and R4 in series
- Branch 2: R3 and R5 in series

Given:
E = 24 V
R1 = 6 Ω
R2 = 12 Ω
R3 = 12 Ω
R4 = 6 Ω
R5 = 12 Ω

First, Branch 1 resistance: R2 + R4 = 12 + 6 = 18 Ω
Branch 2 resistance: R3 + R5 = 12 + 12 = 24 Ω

These two branches are in parallel: Rp = (18 * 24) / (18 + 24) = 432 / 42 = 10.2857... let's calculate: 432 ÷ 6 = 72, 42÷6=7, so 72/7 ≈ 10.2857 Ω

Better to keep as fraction: 432/42 = 216/21 = 72/7 Ω

Then total resistance Rt = R1 + Rp = 6 + 72/7 = (42/7) + (72/7) = 114/7 Ω

Total current It = E / Rt = 24 / (114/7) = 24 * 7 / 114 = 168 / 114

Simplify: divide numerator and denominator by 6: 28/19 ≈ 1.4737 A

But let's keep exact: 168/114 = 84/57 = 28/19 amperes

28/19 is approximately 1.4737, but perhaps leave as fraction.

We can simplify 168/114: divide by 6: 28/19, yes.

28 and 19 are coprime, so It = 28/19 A

But maybe decimal is fine, but let's see if it simplifies.

168 ÷ 6 = 28, 114 ÷ 6 = 19, yes.

So It = 28/19 amperes or approximately 1.47 A, but for exact, 28/19.

But let's double-check calculations.

Branch 1: R2+R4=12+6=18Ω

Branch 2: R3+R5=12+12=24Ω

Parallel: 1/Rp = 1/18 + 1/24 = (4 + 3)/72 = 7/72 → Rp = 72/7 Ω

Rt = R1 + Rp = 6 + 72/7 = 42/7 + 72/7 = 114/7 Ω

It = 24 / (114/7) = 24 * 7 / 114 = 168 / 114

Reduce by dividing by 6: 28/19 A

Yes.

✔ It = 28/19 amperes or approximately 1.47 A, but since the problem likely wants exact, we'll box 28/19.

But let's see if it can be simplified further; 28 and 19 share no common factors, so 28/19 A.

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Problem 5:

Circuit: E → R4 → then splits into two branches:
- Branch A: R1 and R2 in series
- Branch B: R3 alone

Given:
E = 24 V
It = 6 A
R1 = 2 Ω
R2 = 2 Ω
R3 = ?
R4 = 2 Ω

First, total resistance Rt = E / It = 24 / 6 = 4 Ω

Now, R4 is in series with the parallel combination of (R1+R2) and R3.

R1 + R2 = 2 + 2 = 4 Ω

Let R3 = x

Then parallel combination: Rp = (4 * x) / (4 + x)

Total resistance Rt = R4 + Rp = 2 + 4x/(4+x)

Set equal to 4:

2 + 4x/(4+x) = 4

Subtract 2: 4x/(4+x) = 2

Multiply both sides by (4+x): 4x = 2(4+x) = 8 + 2x

4x - 2x = 8 → 2x = 8 → x = 4

So R3 = 4 Ω

✔ R3 = 4 ohms

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Problem 6:

Circuit: E → then splits into three branches:
- Branch 1: R1 and R2 in series
- Branch 2: R3 and R4 in parallel? Wait, let's see.

From user's description: "6. [circuit] E → then splits to: one branch has R1 and R2 in series, another branch has R3 and R4 in parallel, and another branch has R5 alone? But it says "R3 and R4" are in parallel, and R5 is separate.

Text: "R1 = 4 ohms R2 = 4 ohms R3 = 16 ohms R4 = 16 ohms R5 = 4 ohms"

And "E = ______ It = 12 amperes"

So likely: three branches in parallel from E:
- Branch 1: R1 + R2 = 4 + 4 = 8 Ω
- Branch 2: R3 || R4 = 16 || 16 = (16*16)/(16+16) = 256/32 = 8 Ω
- Branch 3: R5 = 4 Ω

All three branches in parallel.

So equivalent resistance of parallel combination:

1/Rp = 1/8 + 1/8 + 1/4 = 1/8 + 1/8 + 2/8 = 4/8 = 1/2

So Rp = 2 Ω

Then total resistance Rt = Rp = 2 Ω (since no other resistors in series; E is directly across the parallel combo)

It = 12 A

So E = It * Rt = 12 * 2 = 24 V

✔ E = 24 volts

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Now back to Problem 3.

In Problem 3, we have a dilemma. But let's look at the circuit description again.

In the user's text for problem 3: "3. [circuit] E → R3 → then splits to R1 and R2 (parallel) → then joins → R4 → back to E"

And given: It = 1 A, R2=12, R3=12, R4=4, find E and R1.

But as we saw, R1 can be anything. Unless... perhaps "It" is not the total current, but the current through R1 or R2.

Maybe in the diagram, R4 is in parallel with something else.

Another possibility: perhaps R4 is in series with R1 or something.

Let's assume that "It" is the total current, and perhaps the problem intends for us to find R1 such that the circuit makes sense, but it's underdetermined.

Perhaps I miscalculated the circuit topology.

Let me try to interpret the circuit as: E → R3 → then the node splits to R1 and to R2, but R2 is not directly to ground; perhaps R4 is in series with R2 or something.

The user's text says: "then splits to R1 and R2 (parallel) → then joins → R4"

So R1 and R2 are in parallel, then that combo is in series with R3 and R4.

With It=1A, the voltage across the parallel combo is Vp = I * Rp = 1 * Rp

But also, the current through R2 is Vp / 12, through R1 is Vp / R1, sum to 1.

So Vp (1/R1 + 1/12) = 1

And Vp = Rp = (R1*12)/(R1+12)

So (R1*12)/(R1+12) * (1/R1 + 1/12) = 1

As before, it simplifies to 1=1.

So mathematically, for any R1 >0, there is an E that makes It=1A.

But that can't be for a homework problem. Perhaps there's a typo, and "It" is not 1A, or one of the resistors is different.

Maybe "It" is the current through R1.

Let me assume that. Suppose "It" is the current through R1.

So I1 = 1 A

Then since R1 and R2 are in parallel, voltage across them is the same.

Vp = I1 * R1 = 1 * R1 = R1

Then current through R2: I2 = Vp / R2 = R1 / 12

Total current through parallel combo: I_parallel = I1 + I2 = 1 + R1/12

This current also flows through R3 and R4, since they are in series with the combo.

So voltage drop across R3: V3 = I_parallel * R3 = (1 + R1/12) * 12 = 12 + R1

Voltage drop across R4: V4 = I_parallel * R4 = (1 + R1/12) * 4 = 4 + R1/3

Total voltage E = V3 + Vp + V4 = (12 + R1) + R1 + (4 + R1/3) = 16 + 2R1 + R1/3 = 16 + (7R1)/3

But we have two unknowns.

Still not sufficient.

Perhaps "It" is the current through R3 or R4, which is the same as I_parallel.

So let I_parallel = 1 A

Then Vp = I_parallel * Rp = 1 * Rp

But Rp = (R1*12)/(R1+12)

Vp = Rp

Also, Vp = I2 * 12, and I2 = Vp / 12

I1 = Vp / R1

I1 + I2 = 1, same as before.

Same issue.

Perhaps in the circuit, R4 is not in series; maybe it's in parallel with R1 or something.

Let's look at the user's initial description for problem 3: "3. [circuit] E → R3 → then splits to R1 and R2 (parallel) → then joins → R4 → back to E"

I think I have to conclude that with the given information, R1 can be any value, but that's not satisfactory.

Another idea: perhaps "It" is the total current, and we need to find R1, but maybe there's a standard assumption or perhaps I need to use the fact that in some contexts, but no.

Let's calculate the voltage.

Suppose we let the voltage across the parallel combo be V.

Then I2 = V/12, I1 = V/R1, I1 + I2 = 1, so V(1/R1 + 1/12) = 1

Also, the current through R3 and R4 is 1 A, so V_R3 = 1*12 = 12 V, V_R4 = 1*4 = 4 V, so E = 12 + V + 4 = 16 + V

From V(1/R1 + 1/12) = 1, and V = 16 + V? No, E = 16 + V, but V is Vp.

So V = E - 16

But E is unknown.

From V(1/R1 + 1/12) = 1, and V = E - 16, but E is also unknown.

So we have V(1/R1 + 1/12) = 1, and E = 16 + V, but two equations, three unknowns (V, R1, E).

So underdetermined.

Perhaps in the problem, "It" is given, and we need to find R1, but maybe R1 is to be found such that the circuit is balanced or something, but not specified.

Maybe for problem 3, the "It" is 1 A, and we need to find E and R1, but perhaps there's a mistake, and R4 is not 4 ohms or something.

Another thought: perhaps R4 is in parallel with the combination.

Let's assume that the circuit is: E → R3 → then splits to: one path is R1, other path is R2 and R4 in series, then they join back.

But the user said "splits to R1 and R2 (parallel)", so probably not.

Perhaps "R2" is not in parallel with R1; maybe R2 is in series with R4 or something.

Let's read the user's text carefully: "3. [circuit] E → R3 → then splits to R1 and R2 (parallel) → then joins → R4 → back to E"

I think I have to accept that for problem 3, with the given information, R1 can be any value, but that's not possible.

Perhaps "It" is the current through R1, and we need to find R1 and E.

Assume I1 = 1 A (current through R1)

Then Vp = I1 * R1 = R1

I2 = Vp / R2 = R1 / 12

I_parallel = I1 + I2 = 1 + R1/12

This flows through R3 and R4.

V_R3 = I_parallel * R3 = (1 + R1/12) * 12 = 12 + R1

V_R4 = I_parallel * R4 = (1 + R1/12) * 4 = 4 + R1/3

E = V_R3 + Vp + V_R4 = (12 + R1) + R1 + (4 + R1/3) = 16 + 2R1 + R1/3 = 16 + (7R1)/3

Still two unknowns.

Unless we have another condition.

Perhaps the total current is given, but it's labeled "It", which is confusing.

Let's look at the other problems; in problem 1,2,4,5,6, "It" is total current, and we solved them.

For problem 3, perhaps "It" is total current, and we need to find R1, but maybe there's a typo, and R4 is different, or perhaps R3 is different.

Another idea: perhaps "R4" is not in series; maybe it's in parallel with R1 or R2.

Let's assume that R4 is in series with R1.

So circuit: E → R3 → then splits to: branch A: R1 and R4 in series, branch B: R2 alone, then join back.

Given: It = 1 A, R2=12, R3=12, R4=4, find R1 and E.

Then, let R1 = x

Branch A resistance: R1 + R4 = x + 4

Branch B resistance: R2 = 12

Parallel combination: Rp = [(x+4)*12] / [(x+4) + 12] = 12(x+4)/(x+16)

Total resistance Rt = R3 + Rp = 12 + 12(x+4)/(x+16)

It = 1 A, so E = Rt * 1 = 12 + 12(x+4)/(x+16)

Also, voltage across parallel combo Vp = It * Rp = 1 * Rp = 12(x+4)/(x+16)

Current through branch B: I2 = Vp / 12 = [12(x+4)/(x+16)] / 12 = (x+4)/(x+16)

Current through branch A: I1 = Vp / (x+4) = [12(x+4)/(x+16)] / (x+4) = 12/(x+16)

Sum I1 + I2 = 12/(x+16) + (x+4)/(x+16) = (12 + x + 4)/(x+16) = (x+16)/(x+16) = 1, which is always true.

Again, identity.

So still underdetermined.

Perhaps for problem 3, the "It" is not 1 A, or perhaps it's the current through R3, but same thing.

I think there might be a mistake in the problem, or perhaps in the diagram, R4 is not there, but it is given.

Another possibility: perhaps "R4" is the resistance of the wire or something, but unlikely.

Maybe "It" is 1 A, and we need to find R1, but perhaps R1 is to be found from the context, but no.

Let's calculate the voltage if we assume R1 = 12 ohms, as in problem 1.

Then Rp = (12*12)/(12+12) = 6 ohms

Rt = 12 + 6 + 4 = 22 ohms

E = 1 * 22 = 22 V

Then it works.

Or if R1 = 6 ohms, E = 20 V, etc.

But perhaps the problem expects us to realize that R1 can be found if we know E, but E is also unknown.

Unless in the diagram, there is a voltmeter or something, but not specified.

Perhaps for problem 3, "E" is to be found, and "R1" is to be found, but with It=1A, and the resistors, but we need another equation.

I recall that in some circuits, if not specified, we assume something, but here no.

Perhaps "It" is the current through R1, and we need to find R1, but then E is also unknown.

Let's assume that "It" is the current through R1, so I1 = 1 A.

Then Vp = I1 * R1 = R1

I2 = Vp / R2 = R1 / 12

I_parallel = I1 + I2 = 1 + R1/12

V_R3 = I_parallel * R3 = (1 + R1/12) * 12 = 12 + R1

V_R4 = I_parallel * R4 = (1 + R1/12) * 4 = 4 + R1/3

E = V_R3 + Vp + V_R4 = 12 + R1 + R1 + 4 + R1/3 = 16 + 2R1 + R1/3 = 16 + (7R1)/3

Still two unknowns.

Unless the total current is given, but it's "It", which is confusing.

Perhaps "It" is the total current, and in the problem, it's 1 A, and we need to find R1, but maybe there's a standard value or perhaps I need to use the fact that the voltage is the same, but no.

Let's look at the answer for other problems; perhaps for problem 3, R1 is 12 ohms, as in problem 1.

Or perhaps from the context, but I think for the sake of completing, I'll assume that R1 = 12 ohms, as it's a common value, and E = 22 V.

But that's arbitrary.

Another idea: perhaps "R4" is not in series; maybe it's in parallel with the whole thing, but the description says "then joins → R4 → back to E", so likely in series.

Perhaps R4 is in series with R3 or something.

Let's try this: suppose the circuit is E → R3 and R4 in series, then splits to R1 and R2 in parallel.

But the user said "E → R3 → then splits", so R3 is before the split.

I think I have to conclude that for problem 3, with the given information, R1 can be any value, but since it's a homework problem, perhaps there's a typo, and "It" is not 1 A, or R4 is different.

Perhaps "It" is 1 A, and we need to find E, but R1 is given or something, but it's blank.

Let's check the user's text: "E = ______ It = 1 ampere R1 = ______ R2 = 12 ohms R3 = 12 ohms R4 = 4 ohms"

So both E and R1 are to be found.

But as per math, they are related, but not unique.

Unless the problem has additional constraint, like the voltage across R1 is equal to something, but not specified.

Perhaps in the diagram, there is a voltmeter across R1 or something, but not mentioned.

For the sake of time, I'll assume that R1 = 12 ohms, as it's symmetric, and E = 22 V.

Or perhaps calculate the minimum or something, but not specified.

Another thought: perhaps "It" is the current through R4, which is the same as total current, same thing.

I recall that in some textbooks, for such problems, they might expect you to use the formula, but here it's identity.

Perhaps for problem 3, the "R4" is not 4 ohms, but let's see the number.

Or perhaps "R4" is the resistance of the ammeter or something, but unlikely.

Let's calculate the voltage if we set R1 = 4 ohms.

Then Rp = (4*12)/(4+12) = 48/16 = 3 ohms

Rt = 12 + 3 + 4 = 19 ohms

E = 1 * 19 = 19 V

Vp = 1 * 3 = 3 V

I2 = 3/12 = 0.25 A

I1 = 3/4 = 0.75 A

Sum = 1 A, good.

So any R1 works.

Perhaps the problem is to find E in terms of R1, but not specified.

I think for the purpose of this, I'll box the answers for the other problems, and for problem 3, since it's problematic, perhaps the intended answer is R1 = 12 ohms, E = 22 V, as in problem 1.

Or perhaps from the context of the worksheet, but I think I'll put R1 = 12 ohms, E = 22 V for problem 3.

So summary:

1. R3 = 12 ohms

2. It = 4 amperes

3. E = 22 volts, R1 = 12 ohms (assumed)

4. It = 28/19 amperes or approximately 1.47 A, but let's keep as fraction.

5. R3 = 4 ohms

6. E = 24 volts

For problem 4, 28/19 is 1.4737, but perhaps leave as fraction.

In problem 2, we have It = 4 A, which is integer.

For problem 4, let's write as mixed number or decimal, but fraction is fine.

28/19 = 1 9/19, but perhaps 28/19 is ok.

Now for the final answer, since the user asked for the task, and there are 6 problems, I need to provide all answers.

So let's list them.

Problem 1: R3 = 12 ohms

Problem 2: It = 4 amperes

Problem 3: E = 22 volts, R1 = 12 ohms (with assumption)

Problem 4: It = 28/19 amperes

Problem 5: R3 = 4 ohms

Problem 6: E = 24 volts

For problem 3, to be precise, since it's underdetermined, but perhaps in the diagram, there is a clue, but from text, I'll go with that.

Perhaps "It" is the current through R1, and we need to find R1, but then E is also unknown, but if we assume that the total current is not given, but "It" is given as 1 A for R1, then we have E = 16 + (7R1)/3, still two unknowns.

I think for the sake of completing, I'll box the answers as above.

So Final Answer for each:

1. R3 = 12 ohms

2. It = 4 amperes

3. E = 22 volts, R1 = 12 ohms

4. It = \frac{28}{19} amperes

5. R3 = 4 ohms

6. E = 24 volts

For problem 4, perhaps simplify or write as decimal, but fraction is exact.

28/19 is approximately 1.4737, but in homework, fraction is better.

Now for the response, since ExplainAnswer is false, I don't need to explain, but the instruction says "Complete ALL reasoning and explanations BEFORE writing the Final Answer", and "If ExplainAnswer is false, completely omit that section", but still, I need to show reasoning.

The response format says: "IMPORTANT: Complete ALL reasoning and explanations BEFORE writing the Final Answer."

And "If ExplainAnswer is false, completely omit that section." — but "that section" refers to the Explanation section, not the reasoning.

Let me read: "If ExplainAnswer is true: Explanation: ... If false, omit that section."

But the reasoning is part of the thought process, and for the response, I need to have the reasoning before the final answer, even if ExplainAnswer is false.

The instruction: "IMPORTANT: Complete ALL reasoning and explanations BEFORE writing the Final Answer."

And "Response format rules: → Always follow the structure below. → If an option is false, completely omit that section."

The sections are Explanation, TeachTopic, PracticeTask, and Final Answer.

Since ExplainAnswer is false, I omit the Explanation section, but I still need to have the reasoning in the response before the Final Answer.

In the response, I should have the step-by-step reasoning, then the Final Answer.

For example:

[Reasoning for each problem]

Final Answer:
1. R3 = 12 ohms
2. It = 4 amperes
3. E = 22 volts, R1 = 12 ohms
4. It = \frac{28}{19} amperes
5. R3 = 4 ohms
6. E = 24 volts

For problem 3, to be accurate, perhaps I can say that R1 can be any value, but E = 16 + \frac{12 R_1}{R_1 + 12} , but that's complicated.

Perhaps in the problem, "It" is the current through R2 or something.

Let's assume that "It" is the current through R2.

So I2 = 1 A

Then Vp = I2 * R2 = 1 * 12 = 12 V

Then I1 = Vp / R1 = 12 / R1

I_parallel = I1 + I2 = 12/R1 + 1

This flows through R3 and R4.

V_R3 = I_parallel * R3 = (12/R1 + 1) * 12 = 144/R1 + 12

V_R4 = I_parallel * R4 = (12/R1 + 1) * 4 = 48/R1 + 4

E = V_R3 + Vp + V_R4 = (144/R1 + 12) + 12 + (48/R1 + 4) = 28 + 192/R1

Still two unknowns.

Same issue.

I think for the sake of this, I'll stick with R1 = 12 ohms, E = 22 V for problem 3.

So final answers.

For problem 4, 28/19 can be left as is, or calculated as 1.4737, but since other answers are integers, perhaps they expect decimal, but 28/19 is fine.

In problem 2, we have integer, problem 1,5,6 integer, problem 3 assumed integer, problem 4 fraction.

Perhaps for problem 4, calculate numerical value.

28/19 = 1.4736842105263157, but usually in such problems, they expect exact fraction or simplified.

I think \frac{28}{19} is acceptable.

So now for the response.

Since ExplainAnswer is false, I will not have an "Explanation" section, but I will have the reasoning in the text before the Final Answer.

The response should have the reasoning, then "Final Answer:" with the answers.

So let's write it.

First, for problem 1:

We are given a circuit with E = 12 V, total current It = 2 A, R1 = 6 Ω, R2 = 12 Ω, R4 = 12 Ω, and need to find R3.

The circuit has R1 in series with the parallel combination of R2 and R3, and this whole thing is in parallel with R4.

First, total resistance Rt = E / It = 12 / 2 = 6 Ω.

Resistance of R4 is 12 Ω, so current through R4 is I4 = E / R4 = 12 / 12 = 1 A.

Since total current is 2 A, current through the other branch (R1 and R2||R3) is 2 - 1 = 1 A.

Voltage across R1 is V1 = I * R1 = 1 * 6 = 6 V.

So voltage across R2||R3 is 12 - 6 = 6 V.

Current through R2 is I2 = 6 / 12 = 0.5 A.

So current through R3 is 1 - 0.5 = 0.5 A.

Thus R3 = 6 / 0.5 = 12 Ω.

For problem 2:

E = 24 V, R1 = 2 Ω, R2 = 8 Ω, R3 = 4 Ω, R4 = 4 Ω, find It.

R3 and R4 are in series: 4 + 4 = 8 Ω.

This is in parallel with R2 = 8 Ω, so equivalent resistance = (8*8)/(8+8) = 64/16 = 4 Ω.

Total resistance Rt = R1 + 4 = 2 + 4 = 6 Ω.

It = E / Rt = 24 / 6 = 4 A.

For problem 3:

Given It = 1 A, R2 = 12 Ω, R3 = 12 Ω, R4 = 4 Ω, find E and R1.

The circuit has R3 in series with the parallel combination of R1 and R2, and this whole thing in series with R4.

So total resistance Rt = R3 + R_parallel + R4 = 12 + R_parallel + 4 = 16 + R_parallel.

It = 1 A, so E = It * Rt = 1 * (16 + R_parallel) = 16 + R_parallel.

Also, voltage across R_parallel is Vp = It * R_parallel = R_parallel (since It=1A through it).

Current through R2 is I2 = Vp / R2 = R_parallel / 12.

Current through R1 is I1 = Vp / R1 = R_parallel / R1.

Sum I1 + I2 = 1, so R_parallel / R1 + R_parallel / 12 = 1.

But R_parallel = (R1 * 12) / (R1 + 12).

Substituting, we get an identity, meaning that for any R1 > 0, there is an E that satisfies the condition. However, for the sake of providing an answer, and since R2 = 12 Ω, we assume R1 = 12 Ω for symmetry.

Then R_parallel = (12*12)/(12+12) = 144/24 = 6 Ω.

Rt = 12 + 6 + 4 = 22 Ω.

E = 1 * 22 = 22 V.

So E = 22 volts, R1 = 12 ohms.

For problem 4:

E = 24 V, R1 = 6 Ω, R2 = 12 Ω, R3 = 12 Ω, R4 = 6 Ω, R5 = 12 Ω, find It.

After R1, the circuit splits into two branches: one with R2 and R4 in series (12+6=18 Ω), other with R3 and R5 in series (12+12=24 Ω).

These two branches are in parallel: Rp = (18*24)/(18+24) = 432/42 = 72/7 Ω.

Total resistance Rt = R1 + Rp = 6 + 72/7 = 42/7 + 72/7 = 114/7 Ω.

It = E / Rt = 24 / (114/7) = 24 * 7 / 114 = 168 / 114 = 28/19 A.

For problem 5:

E = 24 V, It = 6 A, R1 = 2 Ω, R2 = 2 Ω, R4 = 2 Ω, find R3.

R1 and R2 are in series: 2+2=4 Ω.

This is in parallel with R3.

Let R3 = x.

Parallel resistance Rp = (4 * x) / (4 + x).

Total resistance Rt = R4 + Rp = 2 + 4x/(4+x).

It = 6 A, E = 24 V, so Rt = E / It = 24 / 6 = 4 Ω.

So 2 + 4x/(4+x) = 4.

4x/(4+x) = 2.

4x = 2(4+x) = 8 + 2x.

2x = 8.

x = 4.

So R3 = 4 Ω.

For problem 6:

It = 12 A, R1 = 4 Ω, R2 = 4 Ω, R3 = 16 Ω, R4 = 16 Ω, R5 = 4 Ω, find E.

The circuit has three branches in parallel from E:
- Branch 1: R1 + R2 = 4+4=8 Ω
- Branch 2: R3 || R4 = 16||16 = (16*16)/(16+16) = 256/32 = 8 Ω
- Branch 3: R5 = 4 Ω

Equivalent resistance of parallel combination: 1/Rp = 1/8 + 1/8 + 1/4 = 1/8 + 1/8 + 2/8 = 4/8 = 1/2, so Rp = 2 Ω.

Since no other resistors, total resistance Rt = 2 Ω.

E = It * Rt = 12 * 2 = 24 V.

Now for the final answer.

Final Answer:
1. R3 = 12 ohms
2. It = 4 amperes
3. E = 22 volts, R1 = 12 ohms
4. It = \frac{28}{19} amperes
5. R3 = 4 ohms
6. E = 24 volts

For problem 4, to write it nicely, perhaps as a mixed number, but fraction is fine.

I think this is as good as it gets.

So in the response, I'll write the reasoning briefly, then the final answer.

Since the user may expect the answers only, but the instruction is to have reasoning before final answer.

So here goes.