Solved PARALLEL CIRCUIT PROBLEMS Use Ohms Law and the rules ... - Free Printable
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Step-by-step solution for: Solved PARALLEL CIRCUIT PROBLEMS Use Ohms Law and the rules ...
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Step-by-step solution for: Solved PARALLEL CIRCUIT PROBLEMS Use Ohms Law and the rules ...
To solve the problems involving parallel circuits, we will use Ohm's Law and the rules for parallel circuits. Let's go through each problem step by step.
Circuit Diagram:
- Two resistors \( R_1 = 10 \Omega \) and \( R_2 = 10 \Omega \) in parallel.
- Find \( R_T \).
Solution:
For resistors in parallel:
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}
\]
Substitute the values:
\[
\frac{1}{R_T} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}
\]
Thus,
\[
R_T = 5 \Omega
\]
Answer:
\[
R_T = 5 \Omega
\]
---
Circuit Diagram:
- Two resistors \( R_1 = 20 \Omega \) and \( R_2 = 20 \Omega \) in parallel.
- Voltage \( V_T = 20 \text{ V} \).
- Find \( R_T \) and \( I_T \).
Solution:
First, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}
\]
Thus,
\[
R_T = 10 \Omega
\]
Next, find \( I_T \) using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{20}{10} = 2 \text{ A}
\]
Answer:
\[
R_T = 10 \Omega, \quad I_T = 2 \text{ A}
\]
---
Circuit Diagram:
- Two resistors \( R_1 = 20 \Omega \) and \( R_2 \) in parallel.
- Total voltage \( V_T = 4 \text{ V} \).
- Total current \( I_T = 4 \text{ A} \).
- Find \( V_1 \), \( I_1 \), \( I_2 \), \( R_T \), and \( R_2 \).
Solution:
Since the resistors are in parallel, the voltage across each resistor is the same:
\[
V_1 = V_2 = V_T = 4 \text{ V}
\]
Using Ohm's Law to find \( I_1 \):
\[
I_1 = \frac{V_1}{R_1} = \frac{4}{20} = 0.2 \text{ A}
\]
The total current is the sum of the currents through each branch:
\[
I_T = I_1 + I_2
\]
\[
4 = 0.2 + I_2 \implies I_2 = 3.8 \text{ A}
\]
Now, find \( R_2 \) using Ohm's Law:
\[
R_2 = \frac{V_2}{I_2} = \frac{4}{3.8} \approx 1.05 \Omega
\]
Finally, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{1.05} \approx 0.05 + 0.952 \approx 1.002
\]
\[
R_T \approx \frac{1}{1.002} \approx 0.998 \Omega \approx 1 \Omega
\]
Answer:
\[
V_1 = 4 \text{ V}, \quad I_1 = 0.2 \text{ A}, \quad I_2 = 3.8 \text{ A}, \quad R_T \approx 1 \Omega, \quad R_2 \approx 1.05 \Omega
\]
---
Circuit Diagram:
- Two resistors \( R_1 = 30 \Omega \) and \( R_2 = 30 \Omega \) in parallel.
- Total current \( I_T = 4 \text{ A} \).
- Find \( R_T \) and \( V_T \).
Solution:
First, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}
\]
Thus,
\[
R_T = 15 \Omega
\]
Next, find \( V_T \) using Ohm's Law:
\[
V_T = I_T \cdot R_T = 4 \cdot 15 = 60 \text{ V}
\]
Answer:
\[
R_T = 15 \Omega, \quad V_T = 60 \text{ V}
\]
---
Circuit Diagram:
- Two resistors \( R_1 = 30 \Omega \) and \( R_2 = 30 \Omega \) in parallel.
- Total current \( I_T = 2 \text{ A} \).
- Find \( R_T \) and \( V_T \).
Solution:
First, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}
\]
Thus,
\[
R_T = 15 \Omega
\]
Next, find \( V_T \) using Ohm's Law:
\[
V_T = I_T \cdot R_T = 2 \cdot 15 = 30 \text{ V}
\]
Answer:
\[
R_T = 15 \Omega, \quad V_T = 30 \text{ V}
\]
---
Circuit Diagram:
- Two resistors \( R_1 = 40 \Omega \) and \( R_2 = 40 \Omega \) in parallel.
- Total voltage \( V_T = 80 \text{ V} \).
- Find \( R_T \), \( I_T \), \( V_1 \), and \( V_2 \).
Solution:
First, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}
\]
Thus,
\[
R_T = 20 \Omega
\]
Next, find \( I_T \) using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{80}{20} = 4 \text{ A}
\]
Since the resistors are in parallel, the voltage across each resistor is the same:
\[
V_1 = V_2 = V_T = 80 \text{ V}
\]
Answer:
\[
R_T = 20 \Omega, \quad I_T = 4 \text{ A}, \quad V_1 = 80 \text{ V}, \quad V_2 = 80 \text{ V}
\]
---
1. \( R_T = 5 \Omega \)
2. \( R_T = 10 \Omega, \quad I_T = 2 \text{ A} \)
3. \( V_1 = 4 \text{ V}, \quad I_1 = 0.2 \text{ A}, \quad I_2 = 3.8 \text{ A}, \quad R_T \approx 1 \Omega, \quad R_2 \approx 1.05 \Omega \)
4. \( R_T = 15 \Omega, \quad V_T = 60 \text{ V} \)
5. \( R_T = 15 \Omega, \quad V_T = 30 \text{ V} \)
6. \( R_T = 20 \Omega, \quad I_T = 4 \text{ A}, \quad V_1 = 80 \text{ V}, \quad V_2 = 80 \text{ V} \)
\boxed{
\begin{aligned}
1. & \quad R_T = 5 \Omega \\
2. & \quad R_T = 10 \Omega, \quad I_T = 2 \text{ A} \\
3. & \quad V_1 = 4 \text{ V}, \quad I_1 = 0.2 \text{ A}, \quad I_2 = 3.8 \text{ A}, \quad R_T \approx 1 \Omega, \quad R_2 \approx 1.05 \Omega \\
4. & \quad R_T = 15 \Omega, \quad V_T = 60 \text{ V} \\
5. & \quad R_T = 15 \Omega, \quad V_T = 30 \text{ V} \\
6. & \quad R_T = 20 \Omega, \quad I_T = 4 \text{ A}, \quad V_1 = 80 \text{ V}, \quad V_2 = 80 \text{ V}
\end{aligned}
}
Problem 1:
Circuit Diagram:
- Two resistors \( R_1 = 10 \Omega \) and \( R_2 = 10 \Omega \) in parallel.
- Find \( R_T \).
Solution:
For resistors in parallel:
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}
\]
Substitute the values:
\[
\frac{1}{R_T} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}
\]
Thus,
\[
R_T = 5 \Omega
\]
Answer:
\[
R_T = 5 \Omega
\]
---
Problem 2:
Circuit Diagram:
- Two resistors \( R_1 = 20 \Omega \) and \( R_2 = 20 \Omega \) in parallel.
- Voltage \( V_T = 20 \text{ V} \).
- Find \( R_T \) and \( I_T \).
Solution:
First, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}
\]
Thus,
\[
R_T = 10 \Omega
\]
Next, find \( I_T \) using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{20}{10} = 2 \text{ A}
\]
Answer:
\[
R_T = 10 \Omega, \quad I_T = 2 \text{ A}
\]
---
Problem 3:
Circuit Diagram:
- Two resistors \( R_1 = 20 \Omega \) and \( R_2 \) in parallel.
- Total voltage \( V_T = 4 \text{ V} \).
- Total current \( I_T = 4 \text{ A} \).
- Find \( V_1 \), \( I_1 \), \( I_2 \), \( R_T \), and \( R_2 \).
Solution:
Since the resistors are in parallel, the voltage across each resistor is the same:
\[
V_1 = V_2 = V_T = 4 \text{ V}
\]
Using Ohm's Law to find \( I_1 \):
\[
I_1 = \frac{V_1}{R_1} = \frac{4}{20} = 0.2 \text{ A}
\]
The total current is the sum of the currents through each branch:
\[
I_T = I_1 + I_2
\]
\[
4 = 0.2 + I_2 \implies I_2 = 3.8 \text{ A}
\]
Now, find \( R_2 \) using Ohm's Law:
\[
R_2 = \frac{V_2}{I_2} = \frac{4}{3.8} \approx 1.05 \Omega
\]
Finally, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{1.05} \approx 0.05 + 0.952 \approx 1.002
\]
\[
R_T \approx \frac{1}{1.002} \approx 0.998 \Omega \approx 1 \Omega
\]
Answer:
\[
V_1 = 4 \text{ V}, \quad I_1 = 0.2 \text{ A}, \quad I_2 = 3.8 \text{ A}, \quad R_T \approx 1 \Omega, \quad R_2 \approx 1.05 \Omega
\]
---
Problem 4:
Circuit Diagram:
- Two resistors \( R_1 = 30 \Omega \) and \( R_2 = 30 \Omega \) in parallel.
- Total current \( I_T = 4 \text{ A} \).
- Find \( R_T \) and \( V_T \).
Solution:
First, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}
\]
Thus,
\[
R_T = 15 \Omega
\]
Next, find \( V_T \) using Ohm's Law:
\[
V_T = I_T \cdot R_T = 4 \cdot 15 = 60 \text{ V}
\]
Answer:
\[
R_T = 15 \Omega, \quad V_T = 60 \text{ V}
\]
---
Problem 5:
Circuit Diagram:
- Two resistors \( R_1 = 30 \Omega \) and \( R_2 = 30 \Omega \) in parallel.
- Total current \( I_T = 2 \text{ A} \).
- Find \( R_T \) and \( V_T \).
Solution:
First, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}
\]
Thus,
\[
R_T = 15 \Omega
\]
Next, find \( V_T \) using Ohm's Law:
\[
V_T = I_T \cdot R_T = 2 \cdot 15 = 30 \text{ V}
\]
Answer:
\[
R_T = 15 \Omega, \quad V_T = 30 \text{ V}
\]
---
Problem 6:
Circuit Diagram:
- Two resistors \( R_1 = 40 \Omega \) and \( R_2 = 40 \Omega \) in parallel.
- Total voltage \( V_T = 80 \text{ V} \).
- Find \( R_T \), \( I_T \), \( V_1 \), and \( V_2 \).
Solution:
First, find \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}
\]
Thus,
\[
R_T = 20 \Omega
\]
Next, find \( I_T \) using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{80}{20} = 4 \text{ A}
\]
Since the resistors are in parallel, the voltage across each resistor is the same:
\[
V_1 = V_2 = V_T = 80 \text{ V}
\]
Answer:
\[
R_T = 20 \Omega, \quad I_T = 4 \text{ A}, \quad V_1 = 80 \text{ V}, \quad V_2 = 80 \text{ V}
\]
---
Final Answers:
1. \( R_T = 5 \Omega \)
2. \( R_T = 10 \Omega, \quad I_T = 2 \text{ A} \)
3. \( V_1 = 4 \text{ V}, \quad I_1 = 0.2 \text{ A}, \quad I_2 = 3.8 \text{ A}, \quad R_T \approx 1 \Omega, \quad R_2 \approx 1.05 \Omega \)
4. \( R_T = 15 \Omega, \quad V_T = 60 \text{ V} \)
5. \( R_T = 15 \Omega, \quad V_T = 30 \text{ V} \)
6. \( R_T = 20 \Omega, \quad I_T = 4 \text{ A}, \quad V_1 = 80 \text{ V}, \quad V_2 = 80 \text{ V} \)
\boxed{
\begin{aligned}
1. & \quad R_T = 5 \Omega \\
2. & \quad R_T = 10 \Omega, \quad I_T = 2 \text{ A} \\
3. & \quad V_1 = 4 \text{ V}, \quad I_1 = 0.2 \text{ A}, \quad I_2 = 3.8 \text{ A}, \quad R_T \approx 1 \Omega, \quad R_2 \approx 1.05 \Omega \\
4. & \quad R_T = 15 \Omega, \quad V_T = 60 \text{ V} \\
5. & \quad R_T = 15 \Omega, \quad V_T = 30 \text{ V} \\
6. & \quad R_T = 20 \Omega, \quad I_T = 4 \text{ A}, \quad V_1 = 80 \text{ V}, \quad V_2 = 80 \text{ V}
\end{aligned}
}
Parent Tip: Review the logic above to help your child master the concept of series circuit problems worksheet answers.