Hand pointing to a circuit diagram on a whiteboard, demonstrating how to calculate voltage, current, and power in a resistor network.

Let's solve the circuit problem step by step. The goal is to determine the voltage across, current through, and power dissipated by each resistor in the given circuit.

---

🔧 Given Circuit:

We have a 18V battery connected in series with resistors arranged in a combination of series and parallel. Here’s the layout:

- A 100Ω resistor in series with the battery.
- Then, a junction (J₁) splits into two paths:
- One path has a 300Ω resistor going straight down.
- Another path goes through a 50Ω, then a 200Ω, then a 250Ω resistor — but note that the 200Ω is in parallel with the 300Ω? Wait — let's clarify.

But wait — from the "1st redraw" shown in the image, we can infer the correct topology.

---

✔ Step 1: Understand the Redrawn Circuit

The "1st redraw" shows:

- 100Ω → J₁
- From J₁:
- One branch: 50Ω + 250Ω = 300Ω
- Other branch: 200Ω
- These two branches are in parallel between J₁ and J₂.
- Then, from J₂, a 150Ω resistor connects back to the negative terminal of the battery.

Also, there’s a 0Ω resistor shown at the bottom — likely just a wire (short), so it doesn’t affect resistance.

So, the correct equivalent circuit is:

```
18V ──[100Ω]──┬───────────────┐
│ │
├─[50Ω]─[250Ω]──┘
│ │
└─[200Ω]────────┘
│
[150Ω]
│
GND
```

Wait — actually, looking at the original diagram, the 150Ω is in the bottom branch, not after J₂.

Let’s reconstruct carefully.

---

🧩 Reconstructing the Original Circuit

From the original sketch:

- Battery: 18V
- Series with 100Ω → then to junction J₁
- From J₁:
- Downward: 300Ω → to junction J₃
- Right: 50Ω → then 200Ω → then 250Ω → to junction J₂
- But also: from J₁ to J₂ via 300Ω? No — wait.

Actually, looking again:

- There's a 300Ω resistor between J₁ and J₃
- There's a 50Ω from J₁ to some node, then 200Ω to J₂, then 250Ω to J₂?
- And 150Ω from J₃ to J₂?

Wait — here's a better interpretation based on the "1st redraw":

In the "1st redraw", the author shows:

- 100Ω → J₁
- From J₁:
- One path: 50Ω + 250Ω = 300Ω
- Other path: 200Ω
- Both go to J₂
- Then from J₂ → 150Ω → back to battery negative

So this suggests:

> The 300Ω in the original is not present — instead, the 50Ω + 250Ω in series form one branch, and the 200Ω forms another branch — both in parallel between J₁ and J₂.

And the 300Ω might be a mistake or mislabeling.

But wait — in the original diagram, there is a 300Ω resistor from J₁ to J₃, and a 150Ω from J₃ to J₂.

Also, there's a 0Ω resistor from J₂ to J₃ — meaning J₂ and J₃ are the same point (shorted).

Ah! That’s key.

---

🔍 Key Insight: 0Ω Resistor

There is a 0Ω resistor (a short) connecting J₂ and J₃.

So, J₂ and J₃ are at the same potential — they are electrically the same node.

Therefore, we can combine them.

Now, let’s redraw the circuit correctly.

---

✔ Corrected Circuit Topology

Let’s label nodes clearly:

- Battery: 18V positive → 100Ω → J₁
- From J₁:
- Path 1: 300Ω → J₃
- Path 2: 50Ω → 200Ω → 250Ω → J₂
- But J₂ and J₃ are connected by a 0Ω (short), so J₂ ≡ J₃
- So, from J₁:
- One branch: 300Ω → J₃
- Another branch: 50Ω + 200Ω + 250Ω = 500Ω → J₂ = J₃
- Then from J₃ (which is J₂): 150Ω → back to battery negative

So the total circuit is:

```
18V ──[100Ω]──┬───────────────┐
│ │
├─[300Ω]─────────┘
│ │
└─[50Ω+200Ω+250Ω=500Ω]──┘
│
[150Ω]
│
GND
```

So now, the two resistors (300Ω and 500Ω) are in parallel, and that parallel combination is in series with the 100Ω and 150Ω.

---

✔ Step-by-Step Solution

Let’s define:

- R₁ = 100Ω (left)
- R₂ = 300Ω (top branch)
- R₃ = 500Ω (bottom branch: 50 + 200 + 250)
- R₄ = 150Ω (right)

---

🔹 Step 1: Combine Parallel Resistors

R₂ and R₃ are in parallel:

\[
R_{\text{parallel}} = \frac{1}{\frac{1}{300} + \frac{1}{500}} = \frac{1}{\frac{5 + 3}{1500}} = \frac{1}{\frac{8}{1500}} = \frac{1500}{8} = 187.5\,\Omega
\]

---

🔹 Step 2: Total Resistance

Total circuit resistance:

\[
R_{\text{total}} = R_1 + R_{\text{parallel}} + R_4 = 100 + 187.5 + 150 = 437.5\,\Omega
\]

---

🔹 Step 3: Total Current from Battery

\[
I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{18}{437.5} = 0.04114\,\text{A} \approx 41.14\,\text{mA}
\]

---

🔹 Step 4: Voltage Drops

#### Across R₁ (100Ω):

\[
V_{100} = I_{\text{total}} \times 100 = 0.04114 \times 100 = 4.114\,\text{V}
\]

#### Across R₄ (150Ω):

\[
V_{150} = I_{\text{total}} \times 150 = 0.04114 \times 150 = 6.171\,\text{V}
\]

#### Voltage across parallel combination:

\[
V_{\text{parallel}} = V_{\text{total}} - V_{100} - V_{150} = 18 - 4.114 - 6.171 = 7.715\,\text{V}
\]

Alternatively:

\[
V_{\text{parallel}} = I_{\text{total}} \times R_{\text{parallel}} = 0.04114 \times 187.5 = 7.715\,\text{V}
\]

✔ Matches.

---

🔹 Step 5: Currents in Parallel Branches

Now, split the current at J₁.

Current through 300Ω:

\[
I_{300} = \frac{V_{\text{parallel}}}{300} = \frac{7.715}{300} = 0.02572\,\text{A} = 25.72\,\text{mA}
\]

Current through 500Ω (i.e., 50+200+250):

\[
I_{500} = \frac{7.715}{500} = 0.01543\,\text{A} = 15.43\,\text{mA}
\]

Check: \(25.72 + 15.43 = 41.15\) mA ≈ total current ✔

---

🔹 Step 6: Voltages Across Each Resistor

Now compute voltages across individual resistors.

#### For 300Ω:

- Voltage: 7.715 V (same as parallel branch)
- Current: 25.72 mA
- Power: \(P = V \times I = 7.715 \times 0.02572 \approx 0.198\,\text{W}\)

#### For 50Ω:

- Current: 15.43 mA
- Voltage: \(V = I \times R = 0.01543 \times 50 = 0.7715\,\text{V}\)
- Power: \(P = I^2 R = (0.01543)^2 \times 50 \approx 0.0119\,\text{W}\)

#### For 200Ω:

- Same current: 15.43 mA
- Voltage: \(0.01543 \times 200 = 3.086\,\text{V}\)
- Power: \(P = I^2 R = (0.01543)^2 \times 200 \approx 0.0476\,\text{W}\)

#### For 250Ω:

- Same current: 15.43 mA
- Voltage: \(0.01543 \times 250 = 3.8575\,\text{V}\)
- Power: \(P = I^2 R = (0.01543)^2 \times 250 \approx 0.0595\,\text{W}\)

Check: 0.7715 + 3.086 + 3.8575 = 7.715 V ✔

#### For 100Ω:

- Voltage: 4.114 V
- Current: 41.14 mA
- Power: \(P = I^2 R = (0.04114)^2 \times 100 \approx 0.169\,\text{W}\)

#### For 150Ω:

- Voltage: 6.171 V
- Current: 41.14 mA
- Power: \(P = I^2 R = (0.04114)^2 \times 150 \approx 0.253\,\text{W}\)

---

✔ Final Table: Summary

| Resistor | Voltage (V) | Current (A) | Power (W) |
|---------|-------------|-------------|-----------|
| 100Ω | 4.114 | 0.04114 | 0.169 |
| 300Ω | 7.715 | 0.02572 | 0.198 |
| 50Ω | 0.7715 | 0.01543 | 0.0119 |
| 200Ω | 3.086 | 0.01543 | 0.0476 |
| 250Ω | 3.8575 | 0.01543 | 0.0595 |
| 150Ω | 6.171 | 0.04114 | 0.253 |

---

🔚 Conclusion

We successfully broke down the circuit using series/parallel simplification, applied Ohm’s Law, and calculated all required values.

Key steps:

1. Recognized the 0Ω resistor shorts J₂ and J₃ → same node.
2. Identified 300Ω and 500Ω (50+200+250) in parallel.
3. Combined resistances step by step.
4. Found total current.
5. Split current in parallel branches.
6. Calculated voltage, current, and power for each resistor.

This method — break it down, build it up — is essential for complex circuits.

Let me know if you want a visual diagram or simulation!