Series-Parallel Circuit Worksheet #1 for educational use, featuring questions on circuit analysis.

Let’s solve each of the six circuit problems one by one using Ohm’s Law and series/parallel resistance rules.

---

## 🔢 Circuit 1

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

Circuit Configuration:
R2 and R3 are in parallel, and that combination is in series with R1.

Step 1: Find equivalent resistance of R2 || R3

Parallel resistance formula:
\[
R_{23} = \frac{R2 \cdot R3}{R2 + R3} = \frac{4 \cdot 4}{4 + 4} = \frac{16}{8} = 2\ \Omega
\]

Step 2: Total resistance R_T = R1 + R23

From Ohm’s Law:
\[
R_T = \frac{E}{I_T} = \frac{12\ V}{3\ A} = 4\ \Omega
\]

So:
\[
R1 + 2 = 4 \Rightarrow R1 = 2\ \Omega
\]

✔ Answer: R1 = 2 ohms

---

## 🔢 Circuit 2

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

Circuit Configuration:
R1 and R2 are in parallel, and that combination is in series with R3.

Step 1: Find R1 || R2

\[
R_{12} = \frac{4 \cdot 4}{4 + 4} = \frac{16}{8} = 2\ \Omega
\]

Step 2: Total resistance R_T = R12 + R3

From Ohm’s Law:
\[
R_T = \frac{12\ V}{3\ A} = 4\ \Omega
\]

So:
\[
2 + R3 = 4 \Rightarrow R3 = 2\ \Omega
\]

✔ Answer: R3 = 2 ohms

---

## 🔢 Circuit 3

Given:
- E = 12 volts
- R1 = 2 ohms
- R2 = 4 ohms
- R3 = 2 ohms
- R4 = 2 ohms

Circuit Configuration:
R3 and R4 are in series → then that combo is in parallel with R2 → then that whole thing is in series with R1.

Step 1: R3 + R4 = 2 + 2 = 4 Ω

Step 2: (R3+R4) || R2 = 4 Ω || 4 Ω

\[
R_{234} = \frac{4 \cdot 4}{4 + 4} = \frac{16}{8} = 2\ \Omega
\]

Step 3: Total R_T = R1 + R234 = 2 + 2 = 4 Ω

Step 4: I_T = E / R_T = 12 V / 4 Ω = 3 A

✔ Answer: I_T = 3 amperes

---

## 🔢 Circuit 4

Given:
- I_T = 3 amperes
- R1 = 4 ohms
- R2 = 4 ohms
- R3 = 8 ohms

Circuit Configuration:
R1 and R2 are in series → then that combo is in parallel with R3.

Step 1: R1 + R2 = 4 + 4 = 8 Ω

Step 2: (R1+R2) || R3 = 8 Ω || 8 Ω

\[
R_{eq} = \frac{8 \cdot 8}{8 + 8} = \frac{64}{16} = 4\ \Omega
\]

Step 3: E = I_T × R_eq = 3 A × 4 Ω = 12 V

✔ Answer: E = 12 volts

---

## 🔢 Circuit 5

Given:
- E = 12 volts
- R1 = 2 ohms
- R2 = 8 ohms
- R3 = 2 ohms
- R4 = 4 ohms
- R5 = 2 ohms

Circuit Configuration:
- R3, R4, R5 are in series → total = 2+4+2 = 8 Ω
- That 8 Ω is in parallel with R2 (8 Ω)
- Then that combo is in series with R1 (2 Ω)

Step 1: R3+R4+R5 = 8 Ω

Step 2: 8 Ω || R2 (8 Ω) = 4 Ω (since 8*8/(8+8)=4)

Step 3: Total R_T = R1 + 4 = 2 + 4 = 6 Ω

Step 4: I_T = E / R_T = 12 V / 6 Ω = 2 A

✔ Answer: I_T = 2 amperes

---

## 🔢 Circuit 6

Given:
- E = 12 volts
- R1 = 12 ohms
- R2 = 6 ohms
- R3 = 6 ohms
- R4 = 2 ohms

Circuit Configuration:
- R1 is alone on one branch.
- R2 and R3 are in series → 6+6=12 Ω
- That 12 Ω is in parallel with R1 (12 Ω) → gives 6 Ω
- Then that 6 Ω is in series with R4 (2 Ω)

Wait — let’s trace carefully:

Actually, looking at the diagram:

- The top branch has R1 (12 Ω)
- The bottom branch has R2 (6 Ω) and R3 (6 Ω) in series → 12 Ω
- So R1 || (R2+R3) = 12 Ω || 12 Ω = 6 Ω
- Then this 6 Ω is in series with R4 (2 Ω) → Total R_T = 6 + 2 = 8 Ω

But wait — is R4 in series with the parallel combo?

Looking again: Current from battery goes through R4 first, then splits to R1 and (R2+R3). So yes — R4 is in series with the parallel combination of R1 and (R2+R3).

Step 1: R2 + R3 = 6 + 6 = 12 Ω

Step 2: R1 || (R2+R3) = 12 || 12 = 6 Ω

Step 3: R_T = R4 + 6 = 2 + 6 = 8 Ω

Step 4: I_T = E / R_T = 12 V / 8 Ω = 1.5 A

✔ Answer: I_T = 1.5 amperes

---

## ✔ Final Answers:

1. R1 = 2 ohms
2. R3 = 2 ohms
3. I_T = 3 amperes
4. E = 12 volts
5. I_T = 2 amperes
6. I_T = 1.5 amperes

---

## 🧠 Key Concepts Used:
- Ohm’s Law: V = I × R
- Series Resistance: R_total = R1 + R2 + ...
- Parallel Resistance: 1/R_total = 1/R1 + 1/R2 + ... or for two: R_total = (R1×R2)/(R1+R2)
- Always simplify step-by-step from innermost branches outward.

Let me know if you’d like diagrams or step-by-step current/voltage breakdowns for any circuit!