Electric circuits worksheet with problems on calculating equivalent resistance, voltage, current, and resistance in series and parallel circuits.

Problem Analysis and Solution

The worksheet involves solving problems related to electric circuits, specifically focusing on calculating equivalent resistances and analyzing current, voltage, and resistance in series and parallel circuits. Let's solve each part step by step.

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#### Question 1: Determine the equivalent (total) resistance for each of the following circuits.

We are given three circuits (a), (b), and (c). We need to calculate the equivalent resistance for each.

##### Circuit (a):
- Resistors are connected in parallel.
- The formula for equivalent resistance in parallel is:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
- Given resistances: \( R_1 = 7 \, \Omega \), \( R_2 = 5 \, \Omega \), \( R_3 = 2 \, \Omega \).
- Calculate:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{7} + \frac{1}{5} + \frac{1}{2}
\]
- Convert to a common denominator (70):
\[
\frac{1}{7} = \frac{10}{70}, \quad \frac{1}{5} = \frac{14}{70}, \quad \frac{1}{2} = \frac{35}{70}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{10}{70} + \frac{14}{70} + \frac{35}{70} = \frac{59}{70}
\]
- Therefore:
\[
R_{\text{eq}} = \frac{70}{59} \approx 1.18 \, \Omega
\]

##### Circuit (b):
- Resistors are connected in series.
- The formula for equivalent resistance in series is:
\[
R_{\text{eq}} = R_1 + R_2
\]
- Given resistances: \( R_1 = 5 \, \Omega \), \( R_2 = 2 \, \Omega \).
- Calculate:
\[
R_{\text{eq}} = 5 + 2 = 7 \, \Omega
\]

##### Circuit (c):
- Resistors are connected in parallel.
- The formula for equivalent resistance in parallel is:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}
\]
- Given resistances: \( R_1 = 7 \, \Omega \), \( R_2 = 5 \, \Omega \).
- Calculate:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{7} + \frac{1}{5}
\]
- Convert to a common denominator (35):
\[
\frac{1}{7} = \frac{5}{35}, \quad \frac{1}{5} = \frac{7}{35}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{5}{35} + \frac{7}{35} = \frac{12}{35}
\]
- Therefore:
\[
R_{\text{eq}} = \frac{35}{12} \approx 2.92 \, \Omega
\]

However, the provided solution seems to have an error. The correct answer should be approximately \( 2.92 \, \Omega \), not \( 14 \, \Omega \).

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#### Question 2: Fill out the table for the circuit diagramed at the right.

The circuit consists of three resistors (\( R_1 \), \( R_2 \), \( R_3 \)) connected in parallel. The voltage across each resistor is the same (6 V).

##### Given Data:
- Voltage (\( V \)) = 6 V
- Currents:
- \( I_1 = 0.6 \, \text{A} \)
- \( I_2 = 0.3 \, \text{A} \)
- \( I_3 = 0.2 \, \text{A} \)

##### Calculate Resistances:
Using Ohm's Law: \( R = \frac{V}{I} \)

1. For \( R_1 \):
\[
R_1 = \frac{V}{I_1} = \frac{6}{0.6} = 10 \, \Omega
\]

2. For \( R_2 \):
\[
R_2 = \frac{V}{I_2} = \frac{6}{0.3} = 20 \, \Omega
\]

3. For \( R_3 \):
\[
R_3 = \frac{V}{I_3} = \frac{6}{0.2} = 30 \, \Omega
\]

##### Total Resistance:
For parallel resistors, the total resistance is calculated as:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
- Substituting values:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{10} + \frac{1}{20} + \frac{1}{30}
\]
- Convert to a common denominator (60):
\[
\frac{1}{10} = \frac{6}{60}, \quad \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{6}{60} + \frac{3}{60} + \frac{2}{60} = \frac{11}{60}
\]
- Therefore:
\[
R_{\text{eq}} = \frac{60}{11} \approx 5.45 \, \Omega
\]

##### Total Current:
The total current is the sum of the individual currents:
\[
I_{\text{total}} = I_1 + I_2 + I_3 = 0.6 + 0.3 + 0.2 = 1.1 \, \text{A}
\]

##### Final Table:
| Circuit Position | Voltage (V) | Current (A) | Resistance (Ω) |
|------------------|-------------|-------------|----------------|
| 1 | 6 | 0.6 | 10.0 |
| 2 | 6 | 0.3 | 20.0 |
| 3 | 6 | 0.2 | 30.0 |
| Total | 6.00 | 1.1 | 5.45 |

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#### Question 3: Fill out the table for the circuit diagramed at the right.

The circuit consists of a combination of series and parallel resistors. We need to calculate the voltage, current, and resistance for each position.

##### Circuit Diagram Analysis:
- \( R_1 \) is in series with the parallel combination of \( R_2 \) and \( R_3 \).
- Given:
- Voltage across \( R_1 \): \( V_1 = 2.7 \, \text{V} \)
- Current through \( R_1 \): \( I_1 = 0.27 \, \text{A} \)
- Resistance of \( R_1 \): \( R_1 = 10 \, \Omega \)
- Voltage across \( R_2 \) and \( R_3 \): \( V_2 = V_3 = 3.3 \, \text{V} \)
- Current through \( R_2 \): \( I_2 = 0.165 \, \text{A} \)
- Current through \( R_3 \): \( I_3 = 0.11 \, \text{A} \)
- Resistance of \( R_2 \): \( R_2 = 20 \, \Omega \)
- Resistance of \( R_3 \): \( R_3 = 30 \, \Omega \)

##### Verify Calculations:
1. For \( R_1 \):
- Voltage: \( V_1 = 2.7 \, \text{V} \)
- Current: \( I_1 = 0.27 \, \text{A} \)
- Resistance: \( R_1 = \frac{V_1}{I_1} = \frac{2.7}{0.27} = 10 \, \Omega \)

2. For \( R_2 \):
- Voltage: \( V_2 = 3.3 \, \text{V} \)
- Current: \( I_2 = 0.165 \, \text{A} \)
- Resistance: \( R_2 = \frac{V_2}{I_2} = \frac{3.3}{0.165} = 20 \, \Omega \)

3. For \( R_3 \):
- Voltage: \( V_3 = 3.3 \, \text{V} \)
- Current: \( I_3 = 0.11 \, \text{A} \)
- Resistance: \( R_3 = \frac{V_3}{I_3} = \frac{3.3}{0.11} = 30 \, \Omega \)

##### Final Table:
| Circuit Position | Voltage (V) | Current (A) | Resistance (Ω) |
|------------------|-------------|-------------|----------------|
| 1 | 2.7 | 0.27 | 10.0 |
| 2 | 3.3 | 0.165 | 20.0 |
| 3 | 3.3 | 0.11 | 30.0 |

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Final Answers:

1.
- (a) \( R_{\text{eq}} = \frac{70}{59} \approx 1.18 \, \Omega \)
- (b) \( R_{\text{eq}} = 7 \, \Omega \)
- (c) \( R_{\text{eq}} = \frac{35}{12} \approx 2.92 \, \Omega \)

2.
| Circuit Position | Voltage (V) | Current (A) | Resistance (Ω) |
|------------------|-------------|-------------|----------------|
| 1 | 6 | 0.6 | 10.0 |
| 2 | 6 | 0.3 | 20.0 |
| 3 | 6 | 0.2 | 30.0 |
| Total | 6.00 | 1.1 | 5.45 |

3.
| Circuit Position | Voltage (V) | Current (A) | Resistance (Ω) |
|------------------|-------------|-------------|----------------|
| 1 | 2.7 | 0.27 | 10.0 |
| 2 | 3.3 | 0.165 | 20.0 |
| 3 | 3.3 | 0.11 | 30.0 |

\boxed{1.18 \, \Omega, 7 \, \Omega, 2.92 \, \Omega}