In this experiment, you will verify the formulae for the total resistance of resistors connected in series and in parallel. The diagram below shows the circ...
Question 1 Report
In this experiment, you will verify the formulae for the total resistance of resistors connected in series and in parallel.
The diagram below shows the circuit used to measure the resistance of a single resistor or a combination of resistors.
Two resistors labelled R₁ and R₂ are provided. The circuit above is used for each measurement. In each case the appropriate resistor or combination of resistors is connected in place of the dashed component box shown in the diagram. The switch S is closed, the ammeter reading I and the voltmeter reading V are recorded, and the resistance is calculated using R = V / I. The switch is opened between each measurement to prevent the resistors from heating.
Four separate measurements are taken. First, R₁ is connected alone in the circuit. The voltmeter reads 2.51 V and the ammeter reads 0.25 A. Second, R₂ is connected alone. The voltmeter reads 2.62 V and the ammeter reads 0.12 A. Third, R₁ and R₂ are connected in series (end-to-end, so the same current flows through both) and placed in the circuit. The voltmeter reads 2.38 V and the ammeter reads 0.075 A. Fourth, R₁ and R₂ are connected in parallel (both connected between the same two points, so the potential difference is the same across each) and placed in the circuit. The voltmeter reads 2.30 V and the ammeter reads 0.32 A.
The same 3.0 V battery is used throughout the experiment. The digital ammeter reads to 0.01 A on the standard range and to 0.001 A on the milliamp range. The digital voltmeter reads to 0.01 V. All connecting leads are kept short and the same leads are used for all four measurements. All connections are checked to ensure they are tight and free from corrosion before each reading is taken.
(a) Calculate the resistance R₁ from the readings when R₁ is connected alone. [1]
(b) Calculate the resistance R₂ from the readings when R₂ is connected alone. [1]
(c) Using the formula for resistors in series, calculate the expected total resistance Rseries = R₁ + R₂. [1]
(d) From the readings for the series combination, calculate the measured total resistance. [1]
(e) Calculate the percentage difference between your expected and measured values of the total series resistance. [2]
(f) Using the formula for resistors in parallel, calculate the expected total resistance Rparallel using 1/Rparallel = 1/R₁ + 1/R₂. [2]
(g) From the readings for the parallel combination, calculate the measured total resistance and compare it with your expected value from part (f). [2]
(h) State one reason why the measured resistance values may differ from the expected values. [1]
Answer Details
(a) Resistance R₁ [1]
When R₁ is connected alone, the voltmeter reads \( V = 2.51 \) V and the ammeter reads \( I = 0.25 \) A. Using Ohm's law:
\[ R_1 = \frac{V}{I} = \frac{2.51}{0.25} = 10.0 \;\Omega \][1]
(b) Resistance R₂ [1]
When R₂ is connected alone, \( V = 2.62 \) V and \( I = 0.12 \) A:
\[ R_2 = \frac{V}{I} = \frac{2.62}{0.12} = 21.8 \;\Omega \][1]
(c) Expected series resistance [1]
For resistors in series, the total resistance is the sum of the individual resistances:
\[ R_{\text{series}} = R_1 + R_2 = 10.0 + 21.8 = 31.8 \;\Omega \]This works because in a series circuit the same current passes through both resistors, so the total p.d. is the sum of the individual p.d.s, and dividing by the common current gives \( R_{\text{total}} = R_1 + R_2 \). [1]
(d) Measured series resistance [1]
From the series combination readings, \( V = 2.38 \) V and \( I = 0.075 \) A:
\[ R_{\text{series (measured)}} = \frac{V}{I} = \frac{2.38}{0.075} = 31.7 \;\Omega \][1]
(e) Percentage difference (series) [2]
The percentage difference compares how far the measured value is from the expected value:
\[ \text{percentage difference} = \frac{|\text{expected} - \text{measured}|}{\text{expected}} \times 100\% \][1] Substituting:
\[ \text{percentage difference} = \frac{|31.8 - 31.7|}{31.8} \times 100\% = \frac{0.1}{31.8} \times 100\% = 0.3\% \][1]
This very small percentage difference shows excellent agreement between the expected and measured series resistance, confirming the series formula.
(f) Expected parallel resistance [2]
For resistors in parallel, the reciprocal formula is used:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \][1] Substituting the values:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{10.0} + \frac{1}{21.8} = 0.1000 + 0.0459 = 0.1459 \]\[ R_{\text{parallel}} = \frac{1}{0.1459} = 6.9 \;\Omega \][1]
The parallel resistance is always less than the smallest individual resistor. This is because providing an additional path for current to flow reduces the overall resistance of the combination.
(g) Measured parallel resistance and comparison [2]
From the parallel combination readings, \( V = 2.30 \) V and \( I = 0.32 \) A:
\[ R_{\text{parallel (measured)}} = \frac{V}{I} = \frac{2.30}{0.32} = 7.2 \;\Omega \][1]
Comparing with the expected value of 6.9 Ω:
\[ \text{percentage difference} = \frac{|6.9 - 7.2|}{6.9} \times 100\% = \frac{0.3}{6.9} \times 100\% = 4.3\% \]The measured value (7.2 Ω) is close to the expected value (6.9 Ω), with a percentage difference of about 4%. This is within reasonable experimental uncertainty and supports the parallel resistance formula. [1]
(h) Source of difference [1]
The connecting leads and crocodile clips have a small resistance of their own. This contact resistance adds to the measured resistance, making it slightly higher than the calculated value. This effect is proportionally larger for the parallel combination (which has a smaller total resistance), explaining why the percentage difference is larger for the parallel measurement (4.3%) than for the series measurement (0.3%). [1]
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