Solved PARALLEL CIRCUIT PROBLEMS Use Ohms Law and the rules ... - Free Printable

To solve the problems in the image, we need to analyze each circuit step by step. Let's go through each problem systematically.

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Problem 15

Circuit: Resistors \( R_1 = 6 \Omega \), \( R_2 = 12 \Omega \), \( R_3 = 4 \Omega \) in series.
- Given: \( I_T = 3 \, \text{A} \)
- Find: \( V_T \) and \( R_T \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 + R_3 = 6 \, \Omega + 12 \, \Omega + 4 \, \Omega = 22 \, \Omega
\]

2. Total Voltage (\( V_T \)):
Using Ohm's Law: \( V_T = I_T \cdot R_T \)
\[
V_T = 3 \, \text{A} \cdot 22 \, \Omega = 66 \, \text{V}
\]

Answer:
\[
\boxed{66 \, \text{V}, 22 \, \Omega}
\]

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Problem 16

Circuit: Resistors \( R_1 = 3 \, \Omega \), \( R_2 = 6 \, \Omega \), \( R_3 = 2 \, \Omega \) in parallel.
- Given: \( I_T = 6 \, \text{A} \)
- Find: \( R_T \), \( I_1 \), \( I_2 \), \( I_3 \), and \( V_T \)

#### Solution:
1. Total Resistance (\( R_T \)):
For resistors in parallel:
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
\[
\frac{1}{R_T} = \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = \frac{2}{6} + \frac{1}{6} + \frac{3}{6} = \frac{6}{6} = 1
\]
\[
R_T = 1 \, \Omega
\]

2. Total Voltage (\( V_T \)):
Using Ohm's Law: \( V_T = I_T \cdot R_T \)
\[
V_T = 6 \, \text{A} \cdot 1 \, \Omega = 6 \, \text{V}
\]

3. Currents through each resistor:
- \( I_1 \):
\[
I_1 = \frac{V_T}{R_1} = \frac{6 \, \text{V}}{3 \, \Omega} = 2 \, \text{A}
\]
- \( I_2 \):
\[
I_2 = \frac{V_T}{R_2} = \frac{6 \, \text{V}}{6 \, \Omega} = 1 \, \text{A}
\]
- \( I_3 \):
\[
I_3 = \frac{V_T}{R_3} = \frac{6 \, \text{V}}{2 \, \Omega} = 3 \, \text{A}
\]

Answer:
\[
\boxed{1 \, \Omega, 6 \, \text{V}, I_1 = 2 \, \text{A}, I_2 = 1 \, \text{A}, I_3 = 3 \, \text{A}}
\]

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Problem 17

Circuit: Resistors \( R_1 = 20 \, \Omega \), \( R_2 = 80 \, \Omega \), \( R_3 = 16 \, \Omega \) in series.
- Given: \( V_T = 16 \, \text{V} \)
- Find: \( I_T \) and \( R_T \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 + R_3 = 20 \, \Omega + 80 \, \Omega + 16 \, \Omega = 116 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law: \( I_T = \frac{V_T}{R_T} \)
\[
I_T = \frac{16 \, \text{V}}{116 \, \Omega} = \frac{4}{29} \, \text{A} \approx 0.138 \, \text{A}
\]

Answer:
\[
\boxed{\frac{4}{29} \, \text{A}, 116 \, \Omega}
\]

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Problem 18

Circuit: Resistors \( R_1 = 10 \, \Omega \), \( R_2 = 40 \, \Omega \), \( R_3 = 8 \, \Omega \), \( R_4 = 4 \, \Omega \) in series.
- Given: \( I_T = 20 \, \text{A} \)
- Find: \( R_T \), \( V_T \), \( V_1 \), \( I_1 \), and \( I_2 \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 + R_3 + R_4 = 10 \, \Omega + 40 \, \Omega + 8 \, \Omega + 4 \, \Omega = 62 \, \Omega
\]

2. Total Voltage (\( V_T \)):
Using Ohm's Law: \( V_T = I_T \cdot R_T \)
\[
V_T = 20 \, \text{A} \cdot 62 \, \Omega = 1240 \, \text{V}
\]

3. Voltage across \( R_1 \) (\( V_1 \)):
\[
V_1 = I_T \cdot R_1 = 20 \, \text{A} \cdot 10 \, \Omega = 200 \, \text{V}
\]

4. Currents (\( I_1 \) and \( I_2 \)):
Since all resistors are in series, the current is the same through each resistor:
\[
I_1 = I_2 = I_T = 20 \, \text{A}
\]

Answer:
\[
\boxed{62 \, \Omega, 1240 \, \text{V}, V_1 = 200 \, \text{V}, I_1 = 20 \, \text{A}, I_2 = 20 \, \text{A}}
\]

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Problem 19

Circuit: Resistors \( R_1 = 12 \, \Omega \), \( R_2 = 24 \, \Omega \), \( R_3 = 8 \, \Omega \) in series.
- Given: \( V_1 = 16 \, \text{V} \)
- Find: \( R_T \) and \( I_T \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = R_1 + R_2 + R_3 = 12 \, \Omega + 24 \, \Omega + 8 \, \Omega = 44 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law: \( I_T = \frac{V_1}{R_1} \) (since \( V_1 \) is the voltage across \( R_1 \))
\[
I_T = \frac{16 \, \text{V}}{12 \, \Omega} = \frac{4}{3} \, \text{A} \approx 1.33 \, \text{A}
\]

Answer:
\[
\boxed{\frac{4}{3} \, \text{A}, 44 \, \Omega}
\]

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Problem 20

Circuit: Resistors \( 12 \, \Omega \), \( 24 \, \Omega \), \( 8 \, \Omega \), \( 24 \, \Omega \) in series.
- Find: \( R_T \)

#### Solution:
1. Total Resistance (\( R_T \)):
\[
R_T = 12 \, \Omega + 24 \, \Omega + 8 \, \Omega + 24 \, \Omega = 68 \, \Omega
\]

Answer:
\[
\boxed{68 \, \Omega}
\]

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Problem 21

Circuit: Resistors \( R_1 = 6 \, \Omega \), \( R_2 = 2 \, \Omega \), \( R_3 = 20 \, \Omega \) in parallel.
- Given: \( V_T = 30 \, \text{V} \)
- Find: \( R_T \) and \( I_T \)

#### Solution:
1. Total Resistance (\( R_T \)):
For resistors in parallel:
\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
\[
\frac{1}{R_T} = \frac{1}{6} + \frac{1}{2} + \frac{1}{20} = \frac{10}{60} + \frac{30}{60} + \frac{3}{60} = \frac{43}{60}
\]
\[
R_T = \frac{60}{43} \, \Omega \approx 1.395 \, \Omega
\]

2. Total Current (\( I_T \)):
Using Ohm's Law: \( I_T = \frac{V_T}{R_T} \)
\[
I_T = \frac{30 \, \text{V}}{\frac{60}{43} \, \Omega} = 30 \cdot \frac{43}{60} = \frac{1290}{60} = 21.5 \, \text{A}
\]

Answer:
\[
\boxed{\frac{60}{43} \, \Omega, 21.5 \, \text{A}}
\]

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Problem 22

Circuit: A voltage source of \( 10 \, \text{V} \) in series with a \( 5 \, \Omega \) resistor and a \( 12 \, \Omega \) resistor.
- Find: \( V_T \) and \( R_T \)

#### Solution:
1. Total Resistance (\( R_T \)):
The resistors are in series:
\[
R_T = 5 \, \Omega + 12 \, \Omega = 17 \, \Omega
\]

2. Total Voltage (\( V_T \)):
The total voltage is the voltage of the source:
\[
V_T = 10 \, \text{V}
\]

Answer:
\[
\boxed{10 \, \text{V}, 17 \, \Omega}
\]

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Final Answers:

1. Problem 15: \(\boxed{66 \, \text{V}, 22 \, \Omega}\)
2. Problem 16: \(\boxed{1 \, \Omega, 6 \, \text{V}, I_1 = 2 \, \text{A}, I_2 = 1 \, \text{A}, I_3 = 3 \, \text{A}}\)
3. Problem 17: \(\boxed{\frac{4}{29} \, \text{A}, 116 \, \Omega}\)
4. Problem 18: \(\boxed{62 \, \Omega, 1240 \, \text{V}, V_1 = 200 \, \text{V}, I_1 = 20 \, \text{A}, I_2 = 20 \, \text{A}}\)
5. Problem 19: \(\boxed{\frac{4}{3} \, \text{A}, 44 \, \Omega}\)
6. Problem 20: \(\boxed{68 \, \Omega}\)
7. Problem 21: \(\boxed{\frac{60}{43} \, \Omega, 21.5 \, \text{A}}\)
8. Problem 22: \(\boxed{10 \, \text{V}, 17 \, \Omega}\)