### Purpose

To investigate the current flow and voltages in series and parallel circuits, and also to use Ohm’s law to calculate equivalent resistances of series and parallel circuits.

### Hypothesis

The calculated equivalent resistances for the series circuits will abide by the equation R_{eq} = R_{1} + R_{2} and for the parallel circuits the value will be similar to 1/R_{eq} = 1/R_{1} + 1/R_{2}. The current flow is expected to be uniform throughout the series circuits, but will be stronger through the smaller resistor in the parallel circuits. The voltages across each resistor should add up to V_{TOT} in the series circuits, and the voltages should be uniform throughout the parallel circuits.

### Labeled Diagrams

See attached sheet.

### Data

Part 1: Series Circuits

Trial | R_{1} (Ω) | R_{2} (Ω) | I (A) | V_{1} (A) | V_{2} (A) | R_{eq} (Ω) | V_{TOT} (V) |

1 | 10 | 10 | 0.0906 | 0.874 | 0.828 | 19.0 | 1.72 |

2 | 10 | 51 | 0.0368 | 0.355 | 1.78 | 58.2 | 2.14 |

3 | 51 | 51 | 0.0204 | 1.01 | 1.07 | 106 | 2.18 |

Power supply: 2.5 V

Part 2: Parallel Circuits

Trial | R_{1} (Ω) | R_{2} (Ω) | I (A) | V_{1} (A) | V_{2} (A) | R_{eq} (Ω) | V_{TOT} (V) |

1 | 51 | 51 | 0.0701 | 1.72 | 1.74 | 26.5 | 1.86 |

2 | 51 | 68 | 0.0576 | 1.68 | 1.75 | 28.8 | 1.66 |

3 | 68 | 68 | 0.0539 | 1.82 | 1.77 | 33.0 | 1.78 |

Power supply: 2.5 V

Part 3: Currents

R_{1} (Ω) | R_{2} (Ω) | I_{1} (A) | I_{2} (A) | |

1 | 10 | 51 | 0.0655 | 0.0656 |

2 | 51 | 68 | 0.0925 | 0.0648 |

Power supply: 5 V

### Graphs

Part 3: Currents

Trial 1:

Average current through 10 Ω: 0.0655 A

Average current through 51 Ω: 0.0656 A

Trial 2:

Average current through 68 Ω: 0.0648 A

Average current through 51 Ω: 0.0925 A

### Questions

Part 1:

1. Examine the results of Part 1. What is the relationship between the three voltage readings: V_{1}, V_{2}, and V_{TOT}?

V_{1} plus V_{2} is about equal to V_{TOT}.

2. Using the measurements you have made above and your knowledge of Ohm’s law, calculate the equivalent resistance (R_{eq}) of the circuit for each of the three series circuits you tested.

See data table.

3. Study the equivalent resistance readings for the series circuits. For each of the three series circuits, compare the experimental results with the resistance calculated using the rule for calculating equivalent resistance outlined in the Theory section. In evaluating your results, consider the tolerance of each resistor by using the minimum and maximum values in your calculations.

The range for the 10 Ω resistor is 9.5 Ω to 10.5 Ω. This gives a theoretical R_{eq} range of 19.0 Ω to 21.0 Ω for trial 1. The calculated value of 19.0 Ω falls just on the edge of this range. For trial 2, the 10 Ω resistor has the same range and the 51 Ω resistor has a range of 48.45 Ω to 53.55 Ω. This gives a theoretical R_{eq} range of 57.95 Ω to 64.05 Ω. The calculated value of 58.2 Ω again falls just on the edge of that range. For trial 3, the theoretical R_{eq} range is 96.9 Ω to 107.1 Ω, and the calculated R_{eq} value of 106 yet again falls within that range.

Part 2:

4. Using the measurements you have made above and your knowledge of Ohm’s law, calculate the equivalent resistance (R_{eq}) of the circuit for each of the three parallel circuits you tested.

See data table.

5. Study the equivalent resistance readings for the parallel circuits. Do your results verify what is expected for R_{eq} from the Theory section?

Yes, the calculated equivalent resistances have about the same value as 1/R_{eq} = 1/R_{1} + 1/R_{2}.

6. Examine the results of Part 2. What do you notice about the relationship between the three voltage readings V_{1}, V_{2}, and V_{TOT} in parallel circuits?

V_{1}, V_{2}, and V_{TOT} are all about equal.

Part 3:

7. What did you discover about the current flow in a series circuit in Part 3?

The current flow through each resistor is equal.

8. What did you discover about the current flow in a parallel circuit in Part 3?

The current flow through each resistor differs.

9. If the two measured currents in your parallel circuit were not the same, which resistor had the large current going through it? Why?

The smaller resistor had the largest current going through it because current prefers to go through the path of least resistance.

### Conclusion

During part 1 of the experiment….procedure (what was done), results, expectations and sources of error.

During part 2 of the experiment….procedure (what was done), results, expectations and sources of error.

During part 3 of the experiment…procedure (what was done), results, expectations and sources of error.

### Equations

R = V / I

R_{eq} = R_{1} + R_{2} + R_{3} + …

1/R_{eq} = 1/R_{1} + 1/R_{2} + 1/R_{3} + …

V_{TOT }= V_{1} + V_{2} + V_{3} + …