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.

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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.

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## 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 meet before going through $ R5 $ to return to the battery.

So actually:
- $ R1 $ and $ R2 $ are in series → call this branch A: $ R_A = R1 + R2 = 6 + 6 = 12\,\Omega $
- $ R3 $ and $ R4 $ are in series → branch B: $ R_B = R3 + R4 = 6 + 4 = 10\,\Omega $
- Branch A and B are in parallel, so total parallel resistance:
$$
\frac{1}{R_{\text{parallel}}} = \frac{1}{12} + \frac{1}{10} = \frac{5 + 6}{60} = \frac{11}{60}
\Rightarrow R_{\text{parallel}} = \frac{60}{11} \approx 5.45\,\Omega
$$
- Then $ R5 = 6\,\Omega $ is in series with the parallel combination.
- So total resistance:
$$
R_T = R_{\text{parallel}} + R5 = \frac{60}{11} + 6 = \frac{60 + 66}{11} = \frac{126}{11} \approx 11.45\,\Omega
$$

Now find total current $ I_T $:
$$
I_T = \frac{E}{R_T} = \frac{24}{126/11} = 24 \times \frac{11}{126} = \frac{264}{126} = \frac{44}{21} \approx 2.095\,\text{A}
$$

But wait — the question asks for $ I_T $? Let's check what's missing.

Wait — the problem says:
> $ E = 24 $ volts
> $ I_T = ? $
> $ R1 = 6\,\Omega $, $ R2 = 6\,\Omega $, etc.

So we need to compute $ I_T $. But based on above, it's $ \boxed{2.095} $ A.

But let's double-check the circuit structure.

Actually, looking again: Is $ R5 $ in series with the whole thing?

From the diagram:
- Battery → splits into two paths:
- Left: $ R1 $ → $ R2 $
- Right: $ R3 $ → $ R4 $
- Then both paths join together and go through $ R5 $ back to battery.

Yes, so $ R5 $ is in series with the parallel combination of $ (R1+R2) $ and $ (R3+R4) $.

So yes, our calculation is correct.

So:
- $ R_{\text{branch A}} = 6 + 6 = 12\,\Omega $
- $ R_{\text{branch B}} = 6 + 4 = 10\,\Omega $
- Parallel: $ R_P = \frac{1}{1/12 + 1/10} = \frac{1}{(5+6)/60} = 60/11 \approx 5.45\,\Omega $
- $ R_T = R_P + R5 = 60/11 + 6 = 126/11 \approx 11.45\,\Omega $
- $ I_T = E / R_T = 24 / (126/11) = 24 × 11 / 126 = 264 / 126 = 44 / 21 ≈ 2.095\,\text{A} $

✔ Answer: $ I_T = \frac{44}{21} \approx 2.095 $ amperes

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## Problem 2

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

Circuit:
- $ R1 $ and $ R4 $ are in series on left branch.
- $ R2 $ and $ R3 $ are in series on right branch.
- Both branches are in parallel.
- Total current $ I_T = 12 $ A flows into the parallel combination.

First, find equivalent resistance of each branch:
- Left: $ R_{L} = R1 + R4 = 8 + 4 = 12\,\Omega $
- Right: $ R_{R} = R2 + R3 = 4 + 4 = 8\,\Omega $

Now, since they're in parallel:
$$
\frac{1}{R_T} = \frac{1}{12} + \frac{1}{8} = \frac{2 + 3}{24} = \frac{5}{24}
\Rightarrow R_T = \frac{24}{5} = 4.8\,\Omega
$$

Now use Ohm’s Law:
$$
E = I_T \times R_T = 12 \times 4.8 = 57.6\,\text{volts}
$$

✔ Answer: $ E = 57.6 $ volts

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## Problem 3

Given:
- $ E = 12 $ volts
- $ I_T = ? $
- $ R1 = 4\,\Omega $
- $ R2 = 4\,\Omega $
- $ R3 = 4\,\Omega $
- $ R4 = 2\,\Omega $

Circuit:
- $ R1 $ is in series with a parallel combination of $ R2 $ and $ R3 $, then that is in series with $ R4 $?
Wait — let's look at the diagram:

Battery → $ R1 $ → splits into $ R2 $ and $ R3 $ (parallel) → then joins → $ R4 $ → back to battery.

So:
- $ R2 $ and $ R3 $ are in parallel: $ R_{23} = \frac{1}{1/4 + 1/4} = \frac{1}{0.5} = 2\,\Omega $
- Then $ R1 $, $ R_{23} $, and $ R4 $ are in series:
$$
R_T = R1 + R_{23} + R4 = 4 + 2 + 2 = 8\,\Omega
$$
- $ I_T = \frac{E}{R_T} = \frac{12}{8} = 1.5\,\text{A} $

✔ Answer: $ I_T = 1.5 $ amperes

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## Problem 4

Given:
- $ E = ? $
- $ I_T = 6 $ amperes
- $ R1 = 8\,\Omega $
- $ R2 = 4\,\Omega $
- $ R3 = 4\,\Omega $
- $ R4 = ? $ ← Wait, no value given? But we need to find $ E $, so maybe $ R4 $ is not needed?

Wait — values:
- $ R1 = 8\,\Omega $
- $ R2 = 4\,\Omega $
- $ R3 = 4\,\Omega $
- $ R4 = ? $ — but it's not listed. Probably typo or missing.

Wait, recheck:
> $ R1 = 8\,\Omega $, $ R2 = 4\,\Omega $, $ R3 = 4\,\Omega $, $ R4 = ? $

But in the diagram:
- $ R1 $ and $ R4 $ are in series on left branch?
- $ R2 $ and $ R3 $ are in series on right branch?
- Then both branches in parallel?

But $ R4 $ is not given. That can't be.

Wait — perhaps $ R4 $ is not labeled? Or maybe $ R4 $ is $ R3 $? No.

Wait — in the list:
> R1 = 8 ohms
> R2 = 4 ohms
> R3 = 4 ohms
> R4 = ?

But the diagram shows four resistors. Maybe $ R4 $ is missing.

Alternatively, perhaps $ R4 $ is the same as $ R3 $? Unlikely.

Wait — perhaps $ R4 $ is not used? No.

Let me assume the circuit is:
- Left branch: $ R1 $ and $ R4 $ in series → but $ R4 $ unknown.
- Right branch: $ R2 $ and $ R3 $ in series → $ 4 + 4 = 8\,\Omega $

But $ R4 $ is not given — so probably there's an error.

Wait — looking at the image text: "R4 = ?" — but it's not specified.

Wait — no, the worksheet says:
> R1 = 8 ohms
> R2 = 4 ohms
> R3 = 4 ohms
> R4 = ?

But in the diagram, there are four resistors: R1, R2, R3, R4.

But only three values given? Wait — no: R1=8, R2=4, R3=4, and R4 is blank.

That can't be. Unless $ R4 $ is meant to be known.

Wait — perhaps $ R4 $ is missing from the list? Or maybe it's implied?

Wait — maybe the problem is asking for $ E $, and $ R4 $ is part of the circuit.

Alternatively, maybe the diagram has:
- $ R1 $ and $ R4 $ in series on left
- $ R2 $ and $ R3 $ in series on right
- But if $ R4 $ is not given, we can't proceed.

Unless... wait — is $ R4 $ actually $ R3 $? No.

Wait — perhaps $ R4 $ is not present? Or maybe it's a typo.

Wait — looking at the text: "R4 = ?" — so likely $ R4 $ is not provided, which means we cannot solve unless it's implied.

But that doesn’t make sense.

Wait — perhaps the diagram shows:
- $ R1 $ and $ R2 $ in series on left
- $ R3 $ and $ R4 $ in series on right
- But $ R4 $ is not given.

But the values are:
- $ R1 = 8 $
- $ R2 = 4 $
- $ R3 = 4 $
- $ R4 = ? $

Still missing.

Wait — maybe $ R4 $ is also 4 ohms? Not stated.

Alternatively, perhaps $ R4 $ is in parallel with $ R3 $? But no.

Wait — maybe $ R4 $ is shorted? Unlikely.

Wait — perhaps I misread. Let's recheck the original:

> 4.
> E = ?
> I_T = 6 amperes
> R1 = 8 ohms
> R2 = 4 ohms
> R3 = 4 ohms
> R4 = ?

No value for $ R4 $. That’s a problem.

But perhaps $ R4 $ is meant to be 4 ohms? Or maybe it's not needed?

Wait — maybe the circuit is:
- $ R1 $ and $ R2 $ in series on left: $ 8 + 4 = 12\,\Omega $
- $ R3 $ and $ R4 $ in series on right: $ 4 + R4 $
- But $ R4 $ unknown.

Then total resistance depends on $ R4 $, so we can't find $ E $ without it.

This suggests either a typo or missing data.

But perhaps $ R4 $ is not present? Or maybe it's zero?

Alternatively, maybe $ R4 $ is in parallel with $ R3 $? But diagram isn't clear.

Wait — let's assume the standard configuration: two branches in parallel.

Left: $ R1 $ and $ R2 $ in series → $ 8 + 4 = 12\,\Omega $

Right: $ R3 $ and $ R4 $ in series → $ 4 + R4 $

But $ R4 $ unknown.

Unless $ R4 $ is also 4 ohms? Possibly.

But it's not stated.

Wait — perhaps $ R4 $ is not a resistor, but a wire? Unlikely.

Alternatively, maybe $ R4 $ is the same as $ R3 $? No.

Wait — perhaps the problem has a typo, and $ R4 $ is supposed to be given.

But in the image, it's written as "R4 = ?", so maybe we are to leave it blank? But that doesn't help.

Wait — perhaps $ R4 $ is in parallel with $ R3 $? But then it would be different.

Alternatively, maybe $ R4 $ is the load, but not labeled.

This seems incomplete.

But let's assume $ R4 $ is missing, and perhaps it's meant to be 4 ohms? Or maybe it's not needed.

Wait — perhaps $ R4 $ is in series with $ R3 $, and $ R4 = 4\,\Omega $, even though not stated.

Maybe it's a typo and $ R4 = 4\,\Omega $.

Assume $ R4 = 4\,\Omega $ for now.

Then:
- Left branch: $ R1 + R2 = 8 + 4 = 12\,\Omega $
- Right branch: $ R3 + R4 = 4 + 4 = 8\,\Omega $
- Parallel: $ \frac{1}{R_T} = \frac{1}{12} + \frac{1}{8} = \frac{5}{24} \Rightarrow R_T = 4.8\,\Omega $
- $ I_T = 6\,\text{A} $
- $ E = I_T \times R_T = 6 \times 4.8 = 28.8\,\text{V} $

But this assumes $ R4 = 4\,\Omega $. Since it's not given, this is speculative.

Alternatively, perhaps $ R4 $ is not in the circuit? But it's drawn.

Wait — maybe the diagram shows:
- $ R1 $ and $ R2 $ in series on left
- $ R3 $ alone on right
- $ R4 $ is in series with $ R3 $? But not labeled.

This is ambiguous.

But given the pattern, likely $ R4 = 4\,\Omega $, so we'll proceed with that.

✔ Answer (assuming $ R4 = 4\,\Omega $): $ E = 28.8 $ volts

But this is uncertain.

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## Problem 5

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

Circuit:
- $ R1 $ is in series with a parallel combination of $ R2 $, $ R3 $, and $ R4 $? No.

Looking at the diagram:
- Battery → $ R1 $ → then splits into three paths: $ R2 $, $ R3 $, $ R4 $? But $ R4 $ is below.

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

But $ R4 $ is shown in series with $ R1 $? No.

Wait — actually:
- $ R1 $ and $ R4 $ are in series on one path?
- Then $ R2 $ and $ R3 $ are in series on another path?
- And both paths are in parallel?

But the diagram shows:
- From battery → $ R1 $ → then splits into $ R2 $ and $ R3 $ → then joins → $ R4 $ → back to battery.

So:
- $ R2 $ and $ R3 $ are in parallel
- $ R1 $ and $ R4 $ are in series with the parallel combination?

Wait — no:
- Battery → $ R1 $ → then splits into $ R2 $ and $ R3 $ → then both paths join → $ R4 $ → back.

So:
- $ R2 $ and $ R3 $ are in parallel
- $ R1 $ and $ R4 $ are in series with the parallel combo?

Wait — the order is:
- $ R1 $ in series with the parallel combo of $ R2 $ and $ R3 $, then that whole thing in series with $ R4 $?

But $ R4 $ is after the junction.

So:
- $ R1 $ → then parallel branch: $ R2 $ and $ R3 $
- Then after both branches recombine, goes through $ R4 $

So:
- $ R2 $ and $ R3 $ are in parallel: $ R_{23} = \frac{1}{1/2 + 1/2} = 1\,\Omega $
- Then $ R1 $, $ R_{23} $, and $ R4 $ are in series:
$$
R_T = R1 + R_{23} + R4 = 1 + 1 + 1 = 3\,\Omega
$$
- $ I_T = \frac{E}{R_T} = \frac{12}{3} = 4\,\text{A} $

✔ Answer: $ I_T = 4 $ amperes

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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:
- $ R1 $ is in series with a parallel combination of $ R2 $, $ R3 $, $ R4 $, $ R5 $?

Wait — from diagram:
- Battery → $ R1 $ → then splits into:
- $ R2 $
- $ R3 $ → $ R4 $ → $ R5 $ in series?
- Wait — no.

Actually:
- After $ R1 $, the circuit splits:
- One path: $ R2 $
- Another path: $ R3 $, $ R4 $, $ R5 $ in series?

But $ R4 $ and $ R5 $ are shown in parallel with $ R3 $? No.

Wait — diagram:
- $ R1 $ in series with the rest.
- Then parallel branches:
- One branch: $ R2 $
- Other branch: $ R3 $, $ R4 $, $ R5 $ in series?

But $ R4 $ and $ R5 $ are shown in parallel with $ R3 $? No.

Wait — actually:
- $ R3 $ is in series with a parallel combination of $ R4 $ and $ R5 $, and that whole thing is in parallel with $ R2 $.

So:
- $ R4 $ and $ R5 $ are in parallel: $ R_{45} = \frac{1}{1/8 + 1/4} = \frac{1}{(1+2)/8} = \frac{8}{3} \approx 2.67\,\Omega $
- Then $ R3 $ and $ R_{45} $ in series: $ R_{345} = 4 + 8/3 = 12/3 + 8/3 = 20/3 \approx 6.67\,\Omega $
- This branch is in parallel with $ R2 = 4\,\Omega $
- So total parallel resistance:
$$
\frac{1}{R_P} = \frac{1}{4} + \frac{1}{20/3} = \frac{1}{4} + \frac{3}{20} = \frac{5 + 3}{20} = \frac{8}{20} = \frac{2}{5}
\Rightarrow R_P = \frac{5}{2} = 2.5\,\Omega
$$
- Now $ R1 $ is in series with $ R_P $, so total resistance:
$$
R_T = R1 + R_P = R1 + 2.5
$$
- $ I_T = 4\,\text{A} $, $ E = 12\,\text{V} $
- $ R_T = \frac{E}{I_T} = \frac{12}{4} = 3\,\Omega $
- So:
$$
R1 + 2.5 = 3 \Rightarrow R1 = 0.5\,\Omega
$$

✔ Answer: $ R1 = 0.5 $ ohms

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✔ Final Answers:

1. $ I_T = \frac{44}{21} \approx 2.095 $ A
2. $ E = 57.6 $ V
3. $ I_T = 1.5 $ A
4. Incomplete due to missing $ R4 $ — assuming $ R4 = 4\,\Omega $, $ E = 28.8 $ V
5. $ I_T = 4 $ A
6. $ R1 = 0.5 $ Ω

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⚠️ Note on Problem 4:

The value of $ R4 $ is missing. If it's not 4Ω, the answer changes. Please verify the diagram or provide $ R4 $'s value.

Let me know if you'd like a visual explanation or redrawn diagrams!