SOLUTION: Kami export parallelcircuitwkst904 - Studypool - Free Printable

You're working on a Parallel Circuit Problems worksheet. Let’s solve each of the 6 circuits step by step, using the rules for parallel circuits:

---

## 🔑 Key Rules for Parallel Circuits:
1. Voltage is the same across all branches:
\( V_{\text{total}} = V_1 = V_2 = V_3 = \dots \)

2. Total current adds up:
\( I_{\text{total}} = I_1 + I_2 + I_3 + \dots \)

3. Total resistance uses reciprocals:
\( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \)
→ Then take reciprocal to get \( R_{\text{eq}} \)

4. Ohm’s Law applies per branch:
\( I = \frac{V}{R} \)

---

Let’s go circuit by circuit.

---

## 🟦 Circuit 1 (Top Left)
- Battery: 24V
- Resistors: \( R_1 = 12\Omega \), \( R_2 = 24\Omega \)

Step 1: Voltage

→ In parallel, voltage is same across both resistors:
✔ \( V_1 = 24V \), \( V_2 = 24V \)

Step 2: Currents

Using Ohm’s Law:
\( I_1 = \frac{V}{R_1} = \frac{24V}{12\Omega} = 2A \)
\( I_2 = \frac{24V}{24\Omega} = 1A \)

✔ \( I_1 = 2A \), \( I_2 = 1A \)

Step 3: Total Current

\( I_{\text{total}} = I_1 + I_2 = 2A + 1A = 3A \)

✔ \( I_T = 3A \)

Step 4: Equivalent Resistance

Use reciprocals:
\( \frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} \)
→ \( R_{eq} = 8\Omega \)

✔ \( R_{eq} = 8\Omega \)

---

## 🟦 Circuit 2 (Top Right) — labeled “4a”
- Battery: 40V? (Assumed from context; label says “4a”, but likely 40V based on other problems)
- Resistors: \( R_1 = 20\Omega \), \( R_2 = 20\Omega \)

Step 1: Voltage

✔ \( V_1 = 40V \), \( V_2 = 40V \)

Step 2: Currents

\( I_1 = \frac{40V}{20\Omega} = 2A \)
\( I_2 = \frac{40V}{20\Omega} = 2A \)

✔ \( I_1 = 2A \), \( I_2 = 2A \)

Step 3: Total Current

\( I_T = 2A + 2A = 4A \)

✔ \( I_T = 4A \)

Step 4: Equivalent Resistance

\( \frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \)
→ \( R_{eq} = 10\Omega \)

✔ \( R_{eq} = 10\Omega \)

---

## 🟦 Circuit 3 (Middle Left)
- Battery: 6V
- Ammeters show: \( I_1 = 2A \), \( I_2 = 1A \)

We need to find \( R_1 \), \( R_2 \), and \( R_{eq} \)

Step 1: Voltage

✔ \( V_1 = 6V \), \( V_2 = 6V \)

Step 2: Resistances (using Ohm’s Law)

\( R_1 = \frac{V}{I_1} = \frac{6V}{2A} = 3\Omega \)
\( R_2 = \frac{6V}{1A} = 6\Omega \)

✔ \( R_1 = 3\Omega \), \( R_2 = 6\Omega \)

Step 3: Total Current

\( I_T = I_1 + I_2 = 2A + 1A = 3A \)

✔ \( I_T = 3A \)

Step 4: Equivalent Resistance

\( \frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \)
→ \( R_{eq} = 2\Omega \)

✔ \( R_{eq} = 2\Omega \)

---

## 🟦 Circuit 4 (Middle Right)
- Battery: 40V
- Resistors: \( R_1 = 5\Omega \), \( R_2 = 20\Omega \), \( R_3 = 4\Omega \)

Step 1: Voltage

✔ \( V_1 = 40V \), \( V_2 = 40V \), \( V_3 = 40V \)

Step 2: Currents

\( I_1 = \frac{40V}{5\Omega} = 8A \)
\( I_2 = \frac{40V}{20\Omega} = 2A \)
\( I_3 = \frac{40V}{4\Omega} = 10A \)

✔ \( I_1 = 8A \), \( I_2 = 2A \), \( I_3 = 10A \)

Step 3: Total Current

\( I_T = 8A + 2A + 10A = 20A \)

✔ \( I_T = 20A \)

Step 4: Equivalent Resistance

\( \frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{20} + \frac{1}{4} \)

Find common denominator (20):
= \( \frac{4}{20} + \frac{1}{20} + \frac{5}{20} = \frac{10}{20} = \frac{1}{2} \)
→ \( R_{eq} = 2\Omega \)

✔ \( R_{eq} = 2\Omega \)

---

## 🟦 Circuit 5 (Bottom Left)
- Battery: 12V
- Total current: 3A (ammeter shows 3A total)
- \( R_1 = 12\Omega \), \( R_2 = ? \)

We need to find \( I_1 \), \( I_2 \), \( R_2 \), \( R_{eq} \)

Step 1: Voltage

✔ \( V_1 = 12V \), \( V_2 = 12V \)

Step 2: Current through R₁

\( I_1 = \frac{V}{R_1} = \frac{12V}{12\Omega} = 1A \)

Step 3: Current through R₂

Total current = 3A → So \( I_2 = I_T - I_1 = 3A - 1A = 2A \)

✔ \( I_1 = 1A \), \( I_2 = 2A \)

Step 4: Find R₂

\( R_2 = \frac{V}{I_2} = \frac{12V}{2A} = 6\Omega \)

✔ \( R_2 = 6\Omega \)

Step 5: Equivalent Resistance

We can calculate from total voltage and total current:
\( R_{eq} = \frac{V}{I_T} = \frac{12V}{3A} = 4\Omega \)

Or via reciprocals:
\( \frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4} \) → \( R_{eq} = 4\Omega \)

✔ \( R_{eq} = 4\Omega \)

---

## 🟦 Circuit 6 (Bottom Right)
- Battery: 30V (label says “30v” near battery)
- Resistors: \( R_1 = 10\Omega \), \( R_2 = 15\Omega \)

Step 1: Voltage

✔ \( V_1 = 30V \), \( V_2 = 30V \)

Step 2: Currents

\( I_1 = \frac{30V}{10\Omega} = 3A \)
\( I_2 = \frac{30V}{15\Omega} = 2A \)

✔ \( I_1 = 3A \), \( I_2 = 2A \)

Step 3: Total Current

\( I_T = 3A + 2A = 5A \)

✔ \( I_T = 5A \)

Step 4: Equivalent Resistance

\( \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{15} \)

LCM of 10 and 15 is 30:
= \( \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \)
→ \( R_{eq} = 6\Omega \)

✔ \( R_{eq} = 6\Omega \)

---

## ✔ Final Answers Summary:

---

Circuit 1

- \( R_{eq} = 8\Omega \)
- \( I_T = 3A \)
- \( V_1 = 24V \), \( V_2 = 24V \)
- \( I_1 = 2A \), \( I_2 = 1A \)

---

Circuit 2 (“4a”)

- \( R_{eq} = 10\Omega \)
- \( I_T = 4A \)
- \( V_1 = 40V \), \( V_2 = 40V \)
- \( I_1 = 2A \), \( I_2 = 2A \)

---

Circuit 3

- \( V_1 = 6V \), \( V_2 = 6V \)
- \( R_1 = 3\Omega \), \( R_2 = 6\Omega \)
- \( I_T = 3A \)
- \( R_{eq} = 2\Omega \)

---

Circuit 4

- \( R_{eq} = 2\Omega \)
- \( I_T = 20A \)
- \( V_1 = 40V \), \( V_2 = 40V \), \( V_3 = 40V \)
- \( I_1 = 8A \), \( I_2 = 2A \), \( I_3 = 10A \)

---

Circuit 5

- \( V_1 = 12V \), \( V_2 = 12V \)
- \( I_1 = 1A \), \( I_2 = 2A \)
- \( R_2 = 6\Omega \)
- \( R_{eq} = 4\Omega \)

---

Circuit 6

- \( V_1 = 30V \), \( V_2 = 30V \)
- \( I_1 = 3A \), \( I_2 = 2A \)
- \( I_T = 5A \)
- \( R_{eq} = 6\Omega \)

---

## 📝 Note on Worksheet Instructions:
The worksheet says:
> “Remember that in a parallel circuit:
> • the current in the branches... (is the same, adds up)” → ✘ This is misleading.
> Actually: Current splits among branches, and total current = sum of branch currents.

> “the voltage drops... (is the same, adds up to) total voltage” → ✘ Also misleading.
> Actually: Voltage is the same across all branches. It does NOT add up — it's equal everywhere.

> “to calculate total resistance, (add, use reciprocals)” → ✔ Correct! Use reciprocals.

So be careful — the worksheet has some wording errors. The correct rules are as above.

---

✔ You can now fill in all blanks on your worksheet with these values!

Let me know if you’d like this in table format or need help drawing the diagrams too!