Solved Series-Parallel Circuit Worksheet #2 E=24 volis R2=6 ... - Free Printable

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

---

Ohm’s Law:

$$
V = I \times R \quad \text{(Voltage = Current × Resistance)}
$$

In series circuits:
- Current is the same through all components.
- Total resistance $ R_T = R_1 + R_2 + \dots $
- Voltage divides across resistors.

In parallel circuits:
- Voltage is the same across all branches.
- Current divides among branches.
- Total resistance: $ \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \dots $

We’ll go one by one.

---

## Problem 1

Given:
- $ E = 24 $ volts
- $ R1 = 6\,\Omega $
- $ R2 = 6\,\Omega $
- $ R3 = 6\,\Omega $
- $ R4 = 4\,\Omega $
- $ R5 = 6\,\Omega $

Circuit Description:
- $ R1 $ and $ R2 $ are in series, forming a branch.
- $ R3 $ and $ R4 $ are in series, forming another branch.
- These two branches are in parallel with each other.
- $ R5 $ is in series with the entire parallel combination.

Wait — let's analyze the diagram carefully:

Looking at the layout:
- The battery (E) connects to:
- One path: $ R1 $ → $ R2 $
- Another path: $ R3 $ → $ R4 $
- Then both paths connect to $ R5 $, which is on the return path.

So actually:
- $ R1 $ and $ R2 $ are in series.
- $ R3 $ and $ R4 $ are in series.
- These two series combinations are in parallel.
- This whole parallel combination is in series with $ R5 $.

So total resistance:
1. $ R_{A} = R1 + R2 = 6 + 6 = 12\,\Omega $
2. $ R_{B} = R3 + R4 = 6 + 4 = 10\,\Omega $
3. $ R_{parallel} = \frac{1}{\frac{1}{12} + \frac{1}{10}} = \frac{1}{\frac{5+6}{60}} = \frac{60}{11} \approx 5.45\,\Omega $
4. $ R_T = R_{parallel} + R5 = 5.45 + 6 = 11.45\,\Omega $

Now find total current:
$$
I_T = \frac{E}{R_T} = \frac{24}{11.45} \approx 2.10 \text{ amperes}
$$

But wait — the question asks for $ I_T $, so:

✔ Answer: $ I_T \approx 2.10 $ A

---

## Problem 2

Given:
- $ I_T = 12 $ amperes
- $ R1 = 8\,\Omega $
- $ R2 = 4\,\Omega $
- $ R3 = 4\,\Omega $
- $ R4 = 4\,\Omega $

Circuit Description:
- $ R1 $ and $ R4 $ are in series on one branch.
- $ R2 $ and $ R3 $ are in series on another branch.
- Both branches are in parallel.
- The total current splits between the two branches.

So:
- Branch 1: $ R1 + R4 = 8 + 4 = 12\,\Omega $
- Branch 2: $ R2 + R3 = 4 + 4 = 8\,\Omega $
- Total voltage $ E $? We need to find it.

Since the branches are in parallel, they have the same voltage.

Let’s denote:
- $ I_1 $: current through branch 1
- $ I_2 $: current through branch 2

Total current: $ I_T = I_1 + I_2 = 12\,A $

Current divides inversely with resistance:
$$
\frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{8}{12} = \frac{2}{3}
$$

So $ I_1 = \frac{2}{5} \times 12 = 4.8\,A $, $ I_2 = 7.2\,A $

Now, voltage across either branch:
- $ E = I_1 \times (R1 + R4) = 4.8 \times 12 = 57.6\,V $
- Check: $ I_2 \times (R2 + R3) = 7.2 \times 8 = 57.6\,V $

✔ Answer: $ E = 57.6 $ volts

---

## Problem 3

Given:
- $ E = 12 $ volts
- $ I_T = 4 $ amperes
- $ R2 = 4\,\Omega $
- $ R3 = 4\,\Omega $
- $ R4 = 2\,\Omega $

Circuit Description:
- $ R1 $ is in series with a parallel combination of $ R2 $, $ R3 $, and $ R4 $?

Wait — look at the diagram:
- $ R1 $ is on the main line.
- Then there’s a junction: $ R2 $, $ R3 $, $ R4 $ — but how?
- Actually, from the diagram:
- $ R2 $ and $ R3 $ are in series? No — $ R2 $ and $ R3 $ appear to be in parallel with $ R4 $? Wait.

Wait — better analysis:

From left to right:
- Battery → $ R1 $ → then splits into:
- Path 1: $ R2 $
- Path 2: $ R3 $ and $ R4 $ in series
- So:
- $ R2 $ in parallel with $ (R3 + R4) $
- That parallel combo is in series with $ R1 $

So:
- $ R3 + R4 = 4 + 2 = 6\,\Omega $
- Parallel with $ R2 = 4\,\Omega $: $ \frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \Rightarrow R_p = 2.4\,\Omega $
- Total resistance: $ R_T = R1 + R_p = R1 + 2.4 $

But we know:
- $ E = 12\,V $
- $ I_T = 4\,A $
- So $ R_T = \frac{E}{I_T} = \frac{12}{4} = 3\,\Omega $

So:
$$
R1 + 2.4 = 3 \Rightarrow R1 = 0.6\,\Omega
$$

✔ Answer: $ R1 = 0.6\,\Omega $

---

## Problem 4

Given:
- $ E = ? $
- $ I_T = 6 $ amperes
- $ R1 = 8\,\Omega $
- $ R2 = 4\,\Omega $
- $ R3 = 4\,\Omega $

Circuit Description:
- $ R1 $ is in series with a parallel combination of $ R2 $ and $ R3 $.

Wait — looking at the diagram:
- Battery → $ R1 $ → then splits into $ R2 $ and $ R3 $ in parallel, then back to battery.

So:
- $ R2 $ and $ R3 $ are in parallel: $ \frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \Rightarrow R_p = 2\,\Omega $
- Total resistance: $ R_T = R1 + R_p = 8 + 2 = 10\,\Omega $
- $ I_T = 6\,A $
- So $ E = I_T \times R_T = 6 \times 10 = 60\,V $

✔ Answer: $ E = 60 $ volts

---

## Problem 5

Given:
- $ E = 12 $ volts
- $ I_T = 1 $ ampere
- $ R1 = 1\,\Omega $
- $ R2 = 2\,\Omega $
- $ R3 = 2\,\Omega $
- $ R4 = 1\,\Omega $

Circuit Description:
- $ R1 $ and $ R4 $ are in series on the main path?
- But then $ R2 $ and $ R3 $ are in parallel?

Wait — based on the diagram:
- Battery → $ R1 $ → then splits into:
- $ R2 $
- $ R3 $
- Then both recombine and go through $ R4 $ to ground.

So:
- $ R2 $ and $ R3 $ are in parallel: $ \frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow R_p = 1\,\Omega $
- $ R1 $, $ R_p $, and $ R4 $ are in series: $ R_T = 1 + 1 + 1 = 3\,\Omega $
- $ I_T = 1\,A $
- $ E = I_T \times R_T = 1 \times 3 = 3\,V $

But wait! Given $ E = 12\,V $, contradiction?

Wait — problem says $ E = 12\,V $, $ I_T = 1\,A $, so $ R_T = 12\,\Omega $

But we just calculated $ R_T = 3\,\Omega $. That can't be.

Wait — perhaps I misread.

Wait — the given values are:
- $ E = 12\,V $
- $ I_T = 1\,A $
- $ R1 = 1\,\Omega $
- $ R2 = 2\,\Omega $
- $ R3 = 2\,\Omega $
- $ R4 = 1\,\Omega $

And the circuit: $ R1 $ in series with parallel $ R2 $ and $ R3 $, then $ R4 $ in series?

Wait — no. Looking again:

Battery → $ R1 $ → $ R2 $ and $ R3 $ in parallel → $ R4 $ → back.

So yes:
- $ R2 \parallel R3 = 1\,\Omega $
- $ R_T = R1 + R_p + R4 = 1 + 1 + 1 = 3\,\Omega $
- $ I_T = 1\,A $
- $ E = I_T \times R_T = 3\,V $

But problem says $ E = 12\,V $, so something is wrong.

Wait — maybe $ R1 $ and $ R4 $ are not in series with the parallel part?

Wait — could $ R4 $ be in parallel with $ R2 $ and $ R3 $? No — diagram shows $ R4 $ after.

Alternatively, maybe $ R4 $ is in parallel with $ R1 $? No.

Wait — perhaps $ R1 $ and $ R4 $ are in parallel, and $ R2 $ and $ R3 $ in series?

No — based on standard interpretation of this type of diagram:

It looks like:
- $ R1 $ is in series with a parallel combination of $ R2 $ and $ R3 $, and that combination is in series with $ R4 $.

But then $ R_T = 1 + 1 + 1 = 3\,\Omega $, $ E = 12\,V $, so $ I_T = 12/3 = 4\,A $, but given $ I_T = 1\,A $, contradiction.

Wait — unless $ R1 $ and $ R4 $ are in parallel?

Wait — let me re-express the circuit.

Perhaps:
- $ R1 $ and $ R4 $ are in parallel, and $ R2 $ and $ R3 $ are in series, and the two are in parallel?

No — diagram shows:

Battery → $ R1 $ → then splits into $ R2 $ and $ R3 $ → then both go to $ R4 $ → back.

So:
- $ R2 $ and $ R3 $ in parallel → $ R_p = 1\,\Omega $
- $ R1 $, $ R_p $, $ R4 $ in series → $ R_T = 1 + 1 + 1 = 3\,\Omega $
- $ E = 12\,V $ → $ I_T = 12 / 3 = 4\,A $

But given $ I_T = 1\,A $, so inconsistency.

Wait — unless $ R4 $ is not in series?

Wait — maybe $ R4 $ is in parallel with $ R1 $? No.

Wait — could it be that $ R1 $ is in series with the entire parallel network of $ R2 $, $ R3 $, and $ R4 $? But $ R4 $ is shown separately.

Wait — perhaps the circuit is:
- $ R1 $ in series with a parallel combination of $ R2 $, $ R3 $, and $ R4 $?

But $ R2 $ and $ R3 $ are in series, and that series combo is in parallel with $ R4 $? Or all three in parallel?

No — diagram shows $ R2 $ and $ R3 $ in series, and that series combo is in parallel with $ R4 $, and the whole thing in series with $ R1 $?

Let’s assume that.

So:
- $ R2 + R3 = 2 + 2 = 4\,\Omega $
- $ R4 = 1\,\Omega $
- So $ R_p = \frac{1}{\frac{1}{4} + \frac{1}{1}} = \frac{1}{1.25} = 0.8\,\Omega $
- $ R_T = R1 + R_p = 1 + 0.8 = 1.8\,\Omega $
- $ I_T = 1\,A $, $ E = 1.8\,V $, but given $ E = 12\,V $ — still doesn’t match.

This suggests either the diagram is different or we’re missing something.

Wait — perhaps $ R4 $ is in parallel with $ R1 $? Let’s try another configuration.

Wait — common mistake: perhaps $ R1 $ and $ R4 $ are in parallel, and $ R2 $ and $ R3 $ are in series, and the two are in parallel?

But that would make two separate paths.

Wait — let’s look at the diagram structure:

Based on typical layout:
- Battery positive → $ R1 $
- From $ R1 $, it splits:
- One path: $ R2 $
- Another path: $ R3 $ and $ R4 $ in series
- Then both paths meet and go back to battery.

So:
- $ R2 $ in parallel with $ (R3 + R4) $
- $ R3 + R4 = 2 + 1 = 3\,\Omega $
- $ R2 = 2\,\Omega $
- $ R_p = \frac{1}{\frac{1}{2} + \frac{1}{3}} = \frac{1}{\frac{5}{6}} = 1.2\,\Omega $
- $ R_T = R1 + R_p = 1 + 1.2 = 2.2\,\Omega $
- $ I_T = 1\,A $ → $ E = 1 \times 2.2 = 2.2\,V $, but given $ E = 12\,V $ — still mismatch.

Wait — now I see: the problem says “$ E = 12 $ volts” and “$ I_T = 1 $ ampere”, so $ R_T = 12\,\Omega $

So we need to find $ R1 $, but $ R1 $ is already given as $ 1\,\Omega $, and others are known.

So if $ R_T = 12\,\Omega $, and $ R1 = 1\,\Omega $, then the rest must be $ 11\,\Omega $

But the parallel combination of $ R2 $, $ R3 $, $ R4 $ must add up to $ 11\,\Omega $ — but they are only $ 2+2+1=5\,\Omega $ max.

Wait — perhaps $ R1 $ is in parallel with $ R4 $? And $ R2 $ and $ R3 $ in series?

Let’s suppose:
- $ R1 $ and $ R4 $ are in parallel: $ \frac{1}{R_p} = \frac{1}{1} + \frac{1}{1} = 2 \Rightarrow R_p = 0.5\,\Omega $
- $ R2 $ and $ R3 $ in series: $ 2+2=4\,\Omega $
- Now, these two are in parallel? No.

Wait — perhaps the circuit is:
- $ R1 $ in series with $ R2 $, and $ R3 $ in series with $ R4 $, and those two are in parallel?

Then:
- Branch 1: $ R1 + R2 = 1 + 2 = 3\,\Omega $
- Branch 2: $ R3 + R4 = 2 + 1 = 3\,\Omega $
- So $ R_p = \frac{1}{\frac{1}{3} + \frac{1}{3}} = 1.5\,\Omega $
- $ R_T = 1.5\,\Omega $
- $ I_T = 1\,A $ → $ E = 1.5\,V $, not 12.

Still not matching.

Wait — perhaps $ R1 $ is in series with a parallel combination of $ R2 $, $ R3 $, $ R4 $, and the total $ R_T = 12\,\Omega $

But $ R1 = 1\,\Omega $, so the parallel part must be $ 11\,\Omega $

But $ R2 = 2\,\Omega $, $ R3 = 2\,\Omega $, $ R4 = 1\,\Omega $

If all three in parallel: $ \frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} + \frac{1}{1} = 2 \Rightarrow R_p = 0.5\,\Omega $

Too small.

If $ R2 $ and $ R3 $ in series ($ 4\,\Omega $), in parallel with $ R4 = 1\,\Omega $: $ R_p = \frac{1}{\frac{1}{4} + 1} = \frac{1}{1.25} = 0.8\,\Omega $

Still too small.

None of these give $ R_T = 12\,\Omega $

Unless $ R1 $ is not $ 1\,\Omega $ — but it is given.

Wait — the question says: “$ R1 = 1\,\Omega $”, and asks for what? The blank is next to $ R1 $, but it's already given.

Wait — no, looking back:

"5.
E = 12 volts
I_T = 1 ohm
R1 = 1 ohm
R2 = 2 ohms
R3 = 2 ohms
R4 = 1 ohm"

Wait — "I_T = 1 ohm"? That can't be. Ohms is resistance.

Ah! Typo!

It says: "I_T = 1 ohm" — but that should be I_T = 1 ampere, and R1 = ?

Yes! Probably a typo in the worksheet.

Look: in Problem 5, it says:

> E = 12 volts
> I_T = 1 ohm ← this must be a typo; likely meant to be "I_T = 1 ampere", and R1 is unknown.

But it says "R1 = 1 ohm" — wait, no:

Actually, it says:

> E = 12 volts
> I_T = 1 ohm
> R1 = 1 ohm

That doesn't make sense.

Wait — perhaps it's:

> E = 12 volts
> I_T = 1 ampere
> R1 = ?
> R2 = 2 ohms
> R3 = 2 ohms
> R4 = 1 ohm

But it says "R1 = 1 ohm" — so likely the blank is for $ R1 $, but it's given?

No — perhaps the value is missing.

Wait — in the image, it might be that the blank is for $ R1 $, and the text says "R1 = 1 ohm" — but that can't be.

Wait — looking at the original image description:

> 5.
> E = 12 volts
> I_T = 1 ohm
> R1 = 1 ohm
> R2 = 2 ohms
> R3 = 2 ohms
> R4 = 1 ohm

This is clearly a typo. "I_T = 1 ohm" is invalid. It should be "I_T = 1 ampere".

And "R1 = 1 ohm" is given, but the blank is next to R1 — so perhaps the blank is for R1, and the value is not given.

Wait — in the image, it might say "R1 = ___ ohms", and the value is missing.

But in your text, you wrote "R1 = 1 ohm" — likely a misread.

Assuming the correct version is:

- E = 12 V
- I_T = 1 A
- R2 = 2 Ω
- R3 = 2 Ω
- R4 = 1 Ω
- Find R1

And the circuit is: R1 in series with a parallel combination of R2, R3, R4.

But R2 and R3 are in series? Or parallel?

From diagram:
- Battery → R1 → then splits into:
- R2
- R3 and R4 in series
- So:
- Branch 1: R2 = 2 Ω
- Branch 2: R3 + R4 = 2 + 1 = 3 Ω
- Parallel: $ \frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \Rightarrow R_p = 1.2\,\Omega $
- Total resistance: $ R_T = R1 + 1.2 $
- $ I_T = 1\,A $, $ E = 12\,V $ → $ R_T = 12\,\Omega $
- So $ R1 = 12 - 1.2 = 10.8\,\Omega $

✔ Answer: $ R1 = 10.8\,\Omega $

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## Problem 6

Given:
- $ E = 12 $ volts
- $ I_T = 4 $ amperes
- $ R1 = ? $
- $ R2 = 4\,\Omega $
- $ R3 = 4\,\Omega $
- $ R4 = 8\,\Omega $
- $ R5 = 4\,\Omega $

Circuit Description:
- $ R1 $ in series with a parallel combination of:
- $ R2 $
- $ R3 $ and $ R4 $ in series, and that in parallel with $ R5 $? Wait.

Looking at the diagram:
- Battery → R1 → then splits into:
- R2
- R3 and R4 in series
- R5
- So R2, (R3+R4), and R5 are in parallel

So:
- R3 + R4 = 4 + 8 = 12 Ω
- R2 = 4 Ω
- R5 = 4 Ω
- So three branches in parallel: 4Ω, 12Ω, 4Ω

$ \frac{1}{R_p} = \frac{1}{4} + \frac{1}{12} + \frac{1}{4} = \frac{3}{12} + \frac{1}{12} + \frac{3}{12} = \frac{7}{12} \Rightarrow R_p = \frac{12}{7} \approx 1.714\,\Omega $

Total resistance: $ R_T = R1 + R_p = R1 + 1.714 $

$ I_T = 4\,A $, $ E = 12\,V $ → $ R_T = \frac{12}{4} = 3\,\Omega $

So:
$$
R1 = 3 - 1.714 = 1.286\,\Omega
$$

Or exactly:
$$
R_p = \frac{12}{7}, \quad R_T = 3, \quad R1 = 3 - \frac{12}{7} = \frac{21 - 12}{7} = \frac{9}{7} \approx 1.286\,\Omega
$$

✔ Answer: $ R1 = \frac{9}{7} \approx 1.29\,\Omega $

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✔ Final Answers:

1. $ I_T = \frac{24}{11.45} \approx 2.10 $ A
2. $ E = 57.6 $ V
3. $ R1 = 0.6\,\Omega $
4. $ E = 60 $ V
5. $ R1 = 10.8\,\Omega $ (assuming typo in I_T unit)
6. $ R1 = \frac{9}{7} \approx 1.29\,\Omega $

Let me know if you want the detailed diagrams or calculations for any specific one!