Worksheet with six parallel circuit problems using Ohm's Law and parallel circuit principles.

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.

Problem 1:

Circuit Diagram:
- Two resistors \( R_1 = 10 \Omega \) and \( R_2 = 10 \Omega \) in parallel.
- Total resistance \( R_T \) is required.

Solution:
For parallel resistors:
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}
\]
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.
- Total voltage \( V_T = 20 \text{ V} \).
- Total resistance \( R_T \) and total current \( I_T \) are required.

Solution:
First, calculate \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}
\]
Thus,
\[
R_T = 10 \Omega
\]

Next, calculate the total current \( I_T \) using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{20 \text{ V}}{10 \Omega} = 2 \text{ A}
\]

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 \( I_T = 4 \text{ A} \).
- Voltages \( V_1 \) and \( V_2 \), currents \( I_1 \) and \( I_2 \), and resistances \( R_T \) and \( R_2 \) are required.

Solution:
Since the resistors are in parallel, the voltage across each resistor is the same as the total voltage:
\[
V_1 = V_2 = V_T = 4 \text{ V}
\]

Calculate \( I_1 \) using Ohm's Law:
\[
I_1 = \frac{V_1}{R_1} = \frac{4 \text{ V}}{20 \Omega} = 0.2 \text{ A}
\]

The total current is the sum of the currents through each branch:
\[
I_T = I_1 + I_2
\]
\[
4 \text{ A} = 0.2 \text{ A} + I_2
\]
\[
I_2 = 4 \text{ A} - 0.2 \text{ A} = 3.8 \text{ A}
\]

Calculate \( R_2 \) using Ohm's Law:
\[
R_2 = \frac{V_2}{I_2} = \frac{4 \text{ V}}{3.8 \text{ A}} \approx 1.05 \Omega
\]

Calculate \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{1.05}
\]
\[
\frac{1}{R_T} = 0.05 + 0.9524 \approx 1.0024
\]
\[
R_T \approx \frac{1}{1.0024} \approx 0.9976 \Omega \approx 1 \Omega
\]

Answer:
\[
V_1 = 4 \text{ V}, \quad I_1 = 0.2 \text{ A}, \quad R_T \approx 1 \Omega, \quad V_2 = 4 \text{ V}, \quad I_2 = 3.8 \text{ A}, \quad R_2 \approx 1.05 \Omega
\]

---

Problem 4:

Circuit Diagram:
- Two resistors \( R_1 = 30 \Omega \) and \( R_2 = 30 \Omega \) in parallel.
- Total current \( I_T = 4 \text{ A} \).
- Total resistance \( R_T \) and total voltage \( V_T \) are required.

Solution:
First, calculate \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}
\]
Thus,
\[
R_T = 15 \Omega
\]

Next, calculate the total voltage \( V_T \) using Ohm's Law:
\[
V_T = I_T \cdot R_T = 4 \text{ A} \cdot 15 \Omega = 60 \text{ V}
\]

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 \( I_T = 2 \text{ A} \).
- Total resistance \( R_T \) and total voltage \( V_T \) are required.

Solution:
First, calculate \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}
\]
Thus,
\[
R_T = 15 \Omega
\]

Next, calculate the total voltage \( V_T \) using Ohm's Law:
\[
V_T = I_T \cdot R_T = 2 \text{ A} \cdot 15 \Omega = 30 \text{ V}
\]

Since the resistors are in parallel, the voltage across each resistor is the same as the total voltage:
\[
V_2 = V_T = 30 \text{ V}
\]

Answer:
\[
R_T = 15 \Omega, \quad V_T = 30 \text{ V}, \quad V_2 = 30 \text{ V}
\]

---

Problem 6:

Circuit Diagram:
- Two resistors \( R_1 = 40 \Omega \) and \( R_2 = 40 \Omega \) in parallel.
- Total voltage \( V_T = 80 \text{ V} \).
- Total resistance \( R_T \), total current \( I_T \), and voltages \( V_1 \) and \( V_2 \) are required.

Solution:
First, calculate \( R_T \):
\[
\frac{1}{R_T} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}
\]
Thus,
\[
R_T = 20 \Omega
\]

Next, calculate the total current \( I_T \) using Ohm's Law:
\[
I_T = \frac{V_T}{R_T} = \frac{80 \text{ V}}{20 \Omega} = 4 \text{ A}
\]

Since the resistors are in parallel, the voltage across each resistor is the same as the total voltage:
\[
V_1 = V_2 = V_T = 80 \text{ V}
\]

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. \( V_1 = 4 \text{ V}, \quad I_1 = 0.2 \text{ A}, \quad R_T \approx 1 \Omega, \quad V_2 = 4 \text{ V}, \quad I_2 = 3.8 \text{ A}, \quad R_2 \approx 1.05 \Omega \)
4. \( R_T = 15 \Omega, \quad V_T = 60 \text{ V} \)
5. \( R_T = 15 \Omega, \quad V_T = 30 \text{ V}, \quad V_2 = 30 \text{ V} \)
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 R_T \approx 1 \Omega, \quad V_2 = 4 \text{ V}, \quad I_2 = 3.8 \text{ A}, \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}, \quad V_2 = 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}
}