Parallel Circuit Worksheet 1 Answer Key - Fill Online, Printable ... - Free Printable

Let's solve each problem in the Series-Parallel Circuit Worksheet #1 step by step, using Ohm’s Law and principles of series and parallel circuits.

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🔧 Key Concepts:

- Ohm’s Law: $ V = I \times R $
- Series Resistors: $ R_{\text{total}} = R_1 + R_2 + \dots $
- Parallel Resistors: $ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots $
- Voltage in Parallel: Same across all branches
- Current in Series: Same through all components
- Total Current (I_T): Sum of branch currents in parallel

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## ✔ Problem 1:

Given:
- E = 12 volts
- I_T = 3 amperes
- R2 = 4 ohms
- R3 = 4 ohms
- Find: R1 = ?

Step 1: Total Resistance (R_total)

Using Ohm’s Law:
$$
R_{\text{total}} = \frac{E}{I_T} = \frac{12}{3} = 4\ \Omega
$$

Step 2: Analyze the circuit

- R2 and R3 are in parallel.
- Their equivalent resistance:
$$
\frac{1}{R_{23}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \Rightarrow R_{23} = 2\ \Omega
$$

- R1 is in series with this combination.
So:
$$
R_{\text{total}} = R1 + R_{23} = R1 + 2
$$

We already found $ R_{\text{total}} = 4\ \Omega $, so:
$$
R1 + 2 = 4 \Rightarrow R1 = 2\ \Omega
$$

✔ Answer: R1 = 2 ohms

---

## ✔ Problem 2:

Given:
- E = 12 volts
- I_T = 3 amperes
- R1 = 4 ohms
- R2 = 4 ohms
- Find: R3 = ?

Step 1: Total Resistance

$$
R_{\text{total}} = \frac{E}{I_T} = \frac{12}{3} = 4\ \Omega
$$

Step 2: Analyze the circuit

- R1 is in series with the parallel combination of R2 and R3.
- Let’s denote the equivalent of R2 || R3 as $ R_{23} $

Then:
$$
R_{\text{total}} = R1 + R_{23} = 4 + R_{23}
$$

But $ R_{\text{total}} = 4 $, so:
$$
4 + R_{23} = 4 \Rightarrow R_{23} = 0
$$

Wait! That can't be — unless R3 is shorted? But that would mean infinite current.

Wait — let’s double-check the circuit layout.

Looking at diagram 2:
- R1 is in series with a parallel branch containing R2 and R3?
No — actually, from the drawing:
- The battery connects to R1, then splits into two paths: one with R2, one with R3.
Wait — no! Look carefully:

Actually, in Problem 2, R1 is on the main line, and R2 and R3 are in parallel with each other and in series with R1?

Wait — no. The drawing shows:
- R1 is vertical, connected directly from top to bottom.
- R2 is on the right, also vertical.
- R3 is horizontal, at the bottom.

Wait — this is not standard. Let's interpret based on common conventions.

Looking at the schematic:

- Battery E → goes up to a node.
- From there, it splits: one path through R1 down to ground, another path through R2 down to ground.
- But R3 is connected between the bottom of R1 and the bottom of R2? No — wait, R3 is shown horizontally at the bottom.

Ah! It appears:
- R1 is in series with the parallel combination of R2 and R3?

No — actually, looking again:
- R1 is in series with the parallel combination of R2 and R3?
But the wire goes from battery → R1 → split to R2 and R3 → back to battery.

Yes — that makes sense.

So:
- R2 and R3 are in parallel.
- This parallel combo is in series with R1.

So total resistance:
$$
R_{\text{total}} = R1 + \left( \frac{1}{R2} + \frac{1}{R3} \right)^{-1}
$$

We know:
- R1 = 4 Ω
- R2 = 4 Ω
- R_T = 4 Ω (from E/I_T = 12/3 = 4)

So:
$$
R_{\text{total}} = R1 + R_{23} = 4 + R_{23} = 4 \Rightarrow R_{23} = 0
$$

That implies $ R_{23} = 0 $, which means either R2 or R3 is zero, but R2 = 4 Ω.

This suggests a contradiction unless R3 is zero, which is not possible.

Wait — maybe I misread the circuit.

Let me re-analyze Problem 2:

From the diagram:
- Battery E → top node
- From top node: one path through R1 to bottom node
- Another path: through R2 to bottom node
- And R3 is connected between the top node and bottom node, horizontally.

Wait — if R3 is connected across the entire voltage source, then R3 is in parallel with the rest.

But R1 and R2 are both in parallel with each other and with R3?

Wait — let's redraw mentally:

- Top terminal of battery → connects to:
- R1 (down)
- R3 (horizontal)
- R2 (to the right, then down?)

Wait — the diagram likely shows:

- R1 is vertical, from top to bottom.
- R2 is vertical, on the right.
- R3 is horizontal, connecting the bottom of R1 to the bottom of R2.

So the circuit is:

- Battery E → splits into two paths:
- One path: through R1 → down → then through R3 → to R2 → back to battery?
No — that doesn’t make sense.

Wait — better interpretation:

The correct configuration for Problem 2 is:

- Battery E → top node
- From top node:
- Down through R1 to bottom node
- Right through R2 to bottom node
- R3 is connected between the top node and bottom node, meaning it's in parallel with the whole circuit?

But then R1 and R2 are in parallel with each other, and R3 is also in parallel with them.

So:
- R1 = 4 Ω
- R2 = 4 Ω
- R3 = ? (unknown)
- All three resistors are in parallel with each other?

But then total resistance:
$$
\frac{1}{R_T} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3}
= \frac{1}{4} + \frac{1}{4} + \frac{1}{R3} = \frac{1}{2} + \frac{1}{R3}
$$

But we know $ R_T = 4\ \Omega $, so:
$$
\frac{1}{4} = \frac{1}{2} + \frac{1}{R3} \Rightarrow \frac{1}{R3} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}
$$

Negative resistance? Impossible.

So my interpretation must be wrong.

Let’s look at standard configurations.

Wait — perhaps R1 is in series with the parallel combination of R2 and R3.

But earlier calculation gave R23 = 0, which is impossible.

Unless R3 is shorted? But that’s not stated.

Wait — let’s try another possibility.

Maybe R1 is in series with R2, and that series combo is in parallel with R3?

But that doesn't match the diagram.

Wait — here's the key:

In Problem 2, the circuit has:
- Battery E
- R1 is in series with the parallel combination of R2 and R3.

But then total resistance is:
$$
R_T = R1 + (R2 || R3)
$$

We know:
- R1 = 4 Ω
- R2 = 4 Ω
- R_T = 4 Ω

So:
$$
4 = 4 + (R2 || R3) \Rightarrow R2 || R3 = 0
$$

Only possible if R3 = 0 or R2 = 0, but R2 = 4 Ω.

Contradiction.

Therefore, the only way this works is if R3 is not in parallel with R2.

Wait — maybe R3 is in series with R1, and R2 is in parallel with the combination?

No — the diagram shows R1 and R2 both vertical, and R3 horizontal.

After careful analysis, the most plausible configuration is:

- R1 and R2 are in parallel
- R3 is in series with the battery
- So R3 is in series with the parallel combination of R1 and R2

But the diagram shows R3 at the bottom, connected across the bottom.

Wait — perhaps R3 is in parallel with the battery, so it's in parallel with the entire load.

But then the total current would be split among R1, R2, and R3.

But then R1 = 4 Ω, R2 = 4 Ω, R3 = ?

And total current I_T = 3 A, E = 12 V → R_T = 4 Ω

So:
$$
\frac{1}{R_T} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} = \frac{1}{4} + \frac{1}{4} + \frac{1}{R3} = \frac{1}{2} + \frac{1}{R3}
$$

Set equal to $ \frac{1}{4} $:
$$
\frac{1}{2} + \frac{1}{R3} = \frac{1}{4} \Rightarrow \frac{1}{R3} = -\frac{1}{4}
$$

Still impossible.

So something is wrong.

Wait — maybe R3 is in series with the battery?

But the diagram shows R3 at the bottom, like a ground connection.

Perhaps the circuit is:

- Battery → R3 → then splits to R1 and R2 → back to battery

So R3 is in series with the parallel combination of R1 and R2.

That makes sense!

So:
- R1 = 4 Ω
- R2 = 4 Ω
- R3 = ?
- R1 || R2 = $ \frac{1}{1/4 + 1/4} = 2\ \Omega $
- Then total resistance: $ R_T = R3 + 2 $
- But $ R_T = 12 / 3 = 4\ \Omega $
- So $ R3 + 2 = 4 \Rightarrow R3 = 2\ \Omega $

✔ Answer: R3 = 2 ohms

Yes! That works.

So the circuit is:
- R3 in series with the parallel combination of R1 and R2.

Even though R3 is drawn horizontally at the bottom, it's in series with the main loop.

So Problem 2: R3 = 2 ohms

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## ✔ Problem 3:

Given:
- E = 12 V
- R1 = 2 Ω
- R2 = 4 Ω
- R3 = 2 Ω
- R4 = 2 Ω
- Find: I_T = ?

Step 1: Analyze circuit

From diagram:
- R1 is in series with the parallel combination of R2 and (R3 + R4)

Because:
- After R1, the circuit splits:
- One path: R2
- Other path: R3 and R4 in series

So:
- R3 + R4 = 2 + 2 = 4 Ω
- R2 || (R3+R4) = $ \frac{1}{1/4 + 1/4} = 2\ \Omega $
- Then total resistance: R_T = R1 + 2 = 2 + 2 = 4 Ω

Now:
$$
I_T = \frac{E}{R_T} = \frac{12}{4} = 3\ \text{amperes}
$$

✔ Answer: I_T = 3 amperes

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## ✔ Problem 4:

Given:
- I_T = 3 A
- R1 = 4 Ω
- R2 = 4 Ω
- R3 = 8 Ω
- Find: E = ?

Step 1: Analyze circuit

From diagram:
- R1 and R2 are in series → R12 = 4 + 4 = 8 Ω
- This combination is in parallel with R3 = 8 Ω
- So total resistance:
$$
\frac{1}{R_T} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \Rightarrow R_T = 4\ \Omega
$$

Now:
$$
E = I_T \times R_T = 3 \times 4 = 12\ \text{volts}
$$

✔ Answer: E = 12 volts

---

## ✔ Problem 5:

Given:
- E = 12 V
- R1 = 2 Ω
- R2 = 8 Ω
- R3 = 2 Ω
- R4 = 4 Ω
- R5 = 2 Ω
- Find: I_T = ?

Step 1: Analyze circuit

From diagram:
- R1 is in series with the parallel combination of:
- R2 (alone)
- R3 + R4 + R5 (in series)

So:
- R3 + R4 + R5 = 2 + 4 + 2 = 8 Ω
- R2 = 8 Ω
- So R2 || (R3+R4+R5) = $ \frac{1}{1/8 + 1/8} = 4\ \Omega $
- Then total resistance: R_T = R1 + 4 = 2 + 4 = 6 Ω
- I_T = E / R_T = 12 / 6 = 2 A

✔ Answer: I_T = 2 amperes

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## ✔ Problem 6:

Given:
- E = 12 V
- I_T = 12 A
- R1 = 12 Ω
- R2 = 6 Ω
- R3 = 6 Ω
- R4 = 2 Ω
- Find: ??? Wait — nothing is missing?

Wait — the question says "Find: ___" but no blank is provided.

Looking at the worksheet, it says:

> 6.
> E = 12 volts
> I_T = 12 amperes
> R1 = 12 ohms
> R2 = 6 ohms
> R3 = 6 ohms
> R4 = 2 ohms
> [Blank]

But the blank is not labeled — probably a typo.

Wait — perhaps the task is to verify consistency?

But let’s check if the given values are consistent.

Step 1: Analyze circuit

From diagram:
- R1 is in series with R4
- R2 is in series with R3
- These two branches are in parallel

So:
- Left branch: R1 + R4 = 12 + 2 = 14 Ω
- Right branch: R2 + R3 = 6 + 6 = 12 Ω
- Equivalent resistance:
$$
\frac{1}{R_T} = \frac{1}{14} + \frac{1}{12} = \frac{6 + 7}{84} = \frac{13}{84} \Rightarrow R_T = \frac{84}{13} \approx 6.46\ \Omega
$$

Then:
$$
I_T = \frac{E}{R_T} = \frac{12}{84/13} = 12 \times \frac{13}{84} = \frac{156}{84} = 1.857\ \text{A}
$$

But given I_T = 12 A — huge discrepancy.

So either:
- The values are inconsistent
- Or we misread the circuit

Wait — perhaps R1 is not in series with R4?

Look at diagram:
- Battery → splits into two paths:
- Left: R1 → R4 → back
- Right: R2 → R3 → back

Yes — so R1 and R4 in series, R2 and R3 in series, both branches in parallel.

But then:
- Left branch: 12 + 2 = 14 Ω
- Right branch: 6 + 6 = 12 Ω
- Total resistance: ~6.46 Ω
- Expected I_T = 12 / 6.46 ≈ 1.86 A

But given I_T = 12 A → impossible unless voltage is higher.

But E = 12 V.

So contradiction.

Unless... R1 is not in series with R4?

Wait — maybe R4 is in parallel with R1?

But diagram shows R4 below R1.

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

No — the diagram shows:
- Left branch: R1 and R4 in series
- Right branch: R2 and R3 in series
- Both branches in parallel

So our analysis is correct.

But given I_T = 12 A, E = 12 V → R_T = 1 Ω

But calculated R_T ≈ 6.46 Ω → impossible.

So either:
- There's a mistake in the problem
- Or the value of R1 is not 12 Ω?

Wait — the label says “R1 = 12 ohms”, but maybe it’s a typo?

Alternatively, perhaps R1 is in parallel with R4?

But no — the wiring shows R1 above R4, same path.

Wait — maybe R1 is in parallel with R2, and R4 is in parallel with R3?

But the diagram shows R1 and R4 in series on left, R2 and R3 in series on right.

So unless the values are wrong, the given I_T = 12 A cannot be correct.

But perhaps the task is to find something else?

Wait — maybe the blank is for something else, like voltage drop or current?

But the worksheet says “R1 = 12 ohms” and lists others — no blank.

Possibility: The blank is missing, but based on the pattern, perhaps they want us to find I_T, but it’s already given.

Alternatively, maybe the value of R1 is not 12 ohms — but the label says it is.

Wait — perhaps the value of R1 is unknown, but it says “R1 = 12 ohms”.

Wait — recheck: “R1 = 12 ohms” is listed.

But I_T = 12 A, E = 12 V → R_T = 1 Ω

But the minimum resistance is 12 Ω (R1), so impossible.

Therefore, the given values are inconsistent.

But perhaps I misread the circuit.

Another possibility: R1 is in parallel with R2, and R4 is in parallel with R3, and all are in parallel?

But diagram shows R1 and R4 in series on left, R2 and R3 in series on right.

So unless the labels are wrong, there's an error.

Alternatively, maybe R1 is not 12 ohms — but the label says it is.

Wait — perhaps the blank is for R1, but it says “R1 = 12 ohms” — so it’s given.

So maybe the task is just to verify?

But since I_T is given, and E is given, we can compute R_T = 1 Ω.

But actual R_T > 12 Ω → impossible.

So either the problem has a typo, or we misinterpret.

Wait — perhaps R1 is in parallel with R4, and R2 is in parallel with R3, and these two combinations are in series?

But diagram doesn't show that.

Given the inconsistency, and since all other problems are solvable, likely this one has a typo.

But let’s suppose instead that R1 is unknown, and we need to find it.

But it says “R1 = 12 ohms” — so not.

Alternatively, perhaps I_T is not 12 A, but it says it is.

Wait — maybe E is not 12 V? But it says E = 12 volts.

So unless the circuit is different...

Wait — perhaps R4 is in parallel with R1 and R2?

No.

After careful thought, I believe there is a typo in Problem 6.

But let’s assume that R1 is not 12 ohms, and the blank is for R1.

But the label says “R1 = 12 ohms” — so it's given.

Perhaps the blank is for voltage across R1 or current through R1?

But the worksheet format suggests a single blank.

Given the confusion, and since all other problems are consistent, I’ll skip this one for now.

But let’s try to compute what R1 should be if I_T = 12 A, E = 12 V → R_T = 1 Ω

We have:
- Left branch: R1 + R4 = R1 + 2
- Right branch: R2 + R3 = 6 + 6 = 12 Ω
- Total resistance:
$$
\frac{1}{R_T} = \frac{1}{R1 + 2} + \frac{1}{12}
$$
Set $ R_T = 1 $:
$$
\frac{1}{R1 + 2} + \frac{1}{12} = 1
\Rightarrow \frac{1}{R1 + 2} = 1 - \frac{1}{12} = \frac{11}{12}
\Rightarrow R1 + 2 = \frac{12}{11} \approx 1.09\ \Omega
\Rightarrow R1 \approx -0.91\ \Omega
$$

Impossible.

So no solution — the values are inconsistent.

Thus, Problem 6 has conflicting data.

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | R1 = 2 ohms |
| 2 | R3 = 2 ohms |
| 3 | I_T = 3 amperes |
| 4 | E = 12 volts |
| 5 | I_T = 2 amperes |
| 6 | Inconsistent data — likely typo |

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📌 Summary:

- Use Ohm’s Law and series/parallel rules.
- Identify series and parallel combinations carefully.
- For Problem 6, the given values contradict each other — cannot satisfy Ohm’s Law.

Let me know if you'd like a visual explanation or help redrawing any circuit!