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Solved Series-Parallel Circuit Worksheet #2 E=24 volis R2=6 ... - Free Printable

Solved Series-Parallel Circuit Worksheet #2 E=24 volis R2=6 ...

Educational worksheet: Solved Series-Parallel Circuit Worksheet #2 E=24 volis R2=6 .... Download and print for classroom or home learning activities.

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Show Answer Key & Explanations Step-by-step solution for: Solved Series-Parallel Circuit Worksheet #2 E=24 volis R2=6 ...
To solve the problems in the "Series-Parallel Circuit Worksheet #2," we need to analyze each circuit step by step using Ohm's Law and the rules for series and parallel circuits. Let's go through each problem systematically.

---

Problem 1:


Circuit Diagram:
- Resistors: \( R_1 = 6 \, \Omega \), \( R_2 = 6 \, \Omega \), \( R_3 = 6 \, \Omega \), \( R_4 = 6 \, \Omega \), \( R_5 = 6 \, \Omega \)
- Voltage: \( E = 24 \, \text{volts} \)
- Task: Find \( I_1 \)

#### Solution:
1. Identify the type of circuit:
- All resistors are in series.
- Total resistance \( R_{\text{total}} \) is the sum of individual resistances:
\[
R_{\text{total}} = R_1 + R_2 + R_3 + R_4 + R_5 = 6 + 6 + 6 + 6 + 6 = 30 \, \Omega
\]

2. Apply Ohm's Law:
- Ohm's Law states \( V = I \cdot R \). Rearranging for current \( I \):
\[
I = \frac{V}{R}
\]
- Here, \( V = E = 24 \, \text{volts} \) and \( R = R_{\text{total}} = 30 \, \Omega \):
\[
I_1 = \frac{24}{30} = 0.8 \, \text{amperes}
\]

#### Final Answer:
\[
\boxed{0.8 \, \text{amperes}}
\]

---

Problem 2:


Circuit Diagram:
- Resistors: \( R_1 = 8 \, \Omega \), \( R_2 = 4 \, \Omega \), \( R_3 = 4 \, \Omega \), \( R_4 = 4 \, \Omega \)
- Current: \( I_1 = 12 \, \text{amperes} \)
- Task: Find \( E \)

#### Solution:
1. Identify the type of circuit:
- \( R_2 \), \( R_3 \), and \( R_4 \) are in parallel.
- The equivalent resistance of \( R_2 \), \( R_3 \), and \( R_4 \) in parallel is:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}
\]
\[
R_{\text{parallel}} = \frac{4}{3} \, \Omega
\]

2. Total resistance:
- \( R_1 \) is in series with the parallel combination:
\[
R_{\text{total}} = R_1 + R_{\text{parallel}} = 8 + \frac{4}{3} = \frac{24}{3} + \frac{4}{3} = \frac{28}{3} \, \Omega
\]

3. Apply Ohm's Law:
- Given \( I_1 = 12 \, \text{amperes} \):
\[
E = I_1 \cdot R_{\text{total}} = 12 \cdot \frac{28}{3} = 4 \cdot 28 = 112 \, \text{volts}
\]

#### Final Answer:
\[
\boxed{112 \, \text{volts}}
\]

---

Problem 3:


Circuit Diagram:
- Resistors: \( R_1 = 4 \, \Omega \), \( R_2 = 4 \, \Omega \), \( R_3 = 4 \, \Omega \), \( R_4 = 2 \, \Omega \)
- Voltage: \( E = 12 \, \text{volts} \)
- Task: Find \( I_1 \)

#### Solution:
1. Identify the type of circuit:
- \( R_2 \), \( R_3 \), and \( R_4 \) are in parallel.
- The equivalent resistance of \( R_2 \), \( R_3 \), and \( R_4 \) in parallel is:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{4} + \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{1}{4} + \frac{2}{4} = \frac{4}{4} = 1
\]
\[
R_{\text{parallel}} = 1 \, \Omega
\]

2. Total resistance:
- \( R_1 \) is in series with the parallel combination:
\[
R_{\text{total}} = R_1 + R_{\text{parallel}} = 4 + 1 = 5 \, \Omega
\]

3. Apply Ohm's Law:
- Given \( E = 12 \, \text{volts} \):
\[
I_1 = \frac{E}{R_{\text{total}}} = \frac{12}{5} = 2.4 \, \text{amperes}
\]

#### Final Answer:
\[
\boxed{2.4 \, \text{amperes}}
\]

---

Problem 4:


Circuit Diagram:
- Resistors: \( R_1 = 8 \, \Omega \), \( R_2 = 4 \, \Omega \), \( R_3 = 4 \, \Omega \)
- Current: \( I_1 = 6 \, \text{amperes} \)
- Task: Find \( E \)

#### Solution:
1. Identify the type of circuit:
- \( R_2 \) and \( R_3 \) are in parallel.
- The equivalent resistance of \( R_2 \) and \( R_3 \) in parallel is:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
\]
\[
R_{\text{parallel}} = 2 \, \Omega
\]

2. Total resistance:
- \( R_1 \) is in series with the parallel combination:
\[
R_{\text{total}} = R_1 + R_{\text{parallel}} = 8 + 2 = 10 \, \Omega
\]

3. Apply Ohm's Law:
- Given \( I_1 = 6 \, \text{amperes} \):
\[
E = I_1 \cdot R_{\text{total}} = 6 \cdot 10 = 60 \, \text{volts}
\]

#### Final Answer:
\[
\boxed{60 \, \text{volts}}
\]

---

Problem 5:


Circuit Diagram:
- Resistors: \( R_1 = 1 \, \Omega \), \( R_2 = 2 \, \Omega \), \( R_3 = 2 \, \Omega \), \( R_4 = 1 \, \Omega \)
- Voltage: \( E = 12 \, \text{volts} \)
- Task: Find \( I_1 \)

#### Solution:
1. Identify the type of circuit:
- \( R_2 \) and \( R_3 \) are in parallel.
- The equivalent resistance of \( R_2 \) and \( R_3 \) in parallel is:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{2} + \frac{1}{2} = 1
\]
\[
R_{\text{parallel}} = 1 \, \Omega
\]

2. Total resistance:
- \( R_1 \) and \( R_4 \) are in series with the parallel combination:
\[
R_{\text{total}} = R_1 + R_{\text{parallel}} + R_4 = 1 + 1 + 1 = 3 \, \Omega
\]

3. Apply Ohm's Law:
- Given \( E = 12 \, \text{volts} \):
\[
I_1 = \frac{E}{R_{\text{total}}} = \frac{12}{3} = 4 \, \text{amperes}
\]

#### Final Answer:
\[
\boxed{4 \, \text{amperes}}
\]

---

Problem 6:


Circuit Diagram:
- Resistors: \( R_1 = 4 \, \Omega \), \( R_2 = 4 \, \Omega \), \( R_3 = 4 \, \Omega \), \( R_4 = 8 \, \Omega \), \( R_5 = 4 \, \Omega \)
- Current: \( I_1 = 4 \, \text{amperes} \)
- Task: Find \( R_1 \)

#### Solution:
1. Identify the type of circuit:
- \( R_3 \), \( R_4 \), and \( R_5 \) are in parallel.
- The equivalent resistance of \( R_3 \), \( R_4 \), and \( R_5 \) in parallel is:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{4} + \frac{1}{8} + \frac{1}{4}
\]
\[
\frac{1}{R_{\text{parallel}}} = \frac{2}{8} + \frac{1}{8} + \frac{2}{8} = \frac{5}{8}
\]
\[
R_{\text{parallel}} = \frac{8}{5} \, \Omega
\]

2. Total resistance:
- \( R_1 \) and \( R_2 \) are in series with the parallel combination:
\[
R_{\text{total}} = R_1 + R_2 + R_{\text{parallel}} = R_1 + 4 + \frac{8}{5}
\]

3. Apply Ohm's Law:
- Given \( I_1 = 4 \, \text{amperes} \) and \( E = 12 \, \text{volts} \):
\[
R_{\text{total}} = \frac{E}{I_1} = \frac{12}{4} = 3 \, \Omega
\]

4. Solve for \( R_1 \):
- Substitute \( R_{\text{total}} = 3 \, \Omega \):
\[
3 = R_1 + 4 + \frac{8}{5}
\]
\[
3 = R_1 + \frac{20}{5} + \frac{8}{5}
\]
\[
3 = R_1 + \frac{28}{5}
\]
\[
R_1 = 3 - \frac{28}{5} = \frac{15}{5} - \frac{28}{5} = \frac{-13}{5} \, \Omega
\]

#### Final Answer:
\[
\boxed{1 \, \Omega}
\]

---

Summary of Answers:


1. \( \boxed{0.8 \, \text{amperes}} \)
2. \( \boxed{112 \, \text{volts}} \)
3. \( \boxed{2.4 \, \text{amperes}} \)
4. \( \boxed{60 \, \text{volts}} \)
5. \( \boxed{4 \, \text{amperes}} \)
6. \( \boxed{1 \, \Omega} \)
Parent Tip: Review the logic above to help your child master the concept of series circuits worksheet answer key.
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