In this experiment, you will investigate how the total resistance of resistors connected in series depends on the number of resistors. You uses identical 10...
Question 1 Report
In this experiment, you will investigate how the total resistance of resistors connected in series depends on the number of resistors. You uses identical 10 Ω resistors. You connects one resistor to a battery, a switch, an ammeter and a voltmeter as shown in Fig. 2.1. You then adds resistors one at a time in series, measuring the potential difference across all the resistors and the current each time.
You records your results in Table 2.1.
| Number of resistors n | V / V | I / A | R / Ω |
|---|---|---|---|
| 1 | 1.42 | 0.14 | |
| 2 | 1.45 | 0.07 | |
| 3 | 1.46 | 0.05 | |
| 4 | 1.47 | 0.04 |
(a) Complete Table 2.1 by calculating R for each row. Give your answers to the nearest whole number. [2]
(b) State the relationship between the total resistance and the number of resistors connected in series. [1]
(c) Use your results to predict the total resistance when five identical 10 Ω resistors are connected in series. [1]
(d) You notices that the voltmeter reading barely changes as more resistors are added. Explain why the current decreases even though the voltage stays almost the same. [2]
(e) Describe one change you could make to the method to obtain more reliable results. [1]
(f) Record the resolution of the ammeter from the readings in the table. [1]
(g) Describe how you should connect the voltmeter in the circuit. [1]
(h) State one source of uncertainty in this experiment. [1]
Answer Details
(a) Using \( R = \frac{V}{I} \) for each row:
| n | V / V | I / A | R / Ω |
|---|---|---|---|
| 1 | 1.42 | 0.14 | 10 |
| 2 | 1.45 | 0.07 | 21 |
| 3 | 1.46 | 0.05 | 29 |
| 4 | 1.47 | 0.04 | 37 |
For example, with 2 resistors: \( R = \frac{1.45}{0.07} = 20.7 \approx 21 \; \Omega \). [1] All values are rounded to the nearest whole number. [1]
(b) The total resistance is approximately proportional to the number of resistors. Adding one 10 Ω resistor increases the total resistance by approximately 10 Ω. [1]
(c) Following the pattern, five 10 Ω resistors in series would give a total resistance of approximately \( 5 \times 10 = 50 \; \Omega \). [1]
(d) The battery maintains a nearly constant e.m.f., so the voltage across the resistors stays almost the same as more are added. [1] Since \( I = \frac{V}{R} \), increasing the total resistance while the voltage remains roughly constant means the current must decrease. [1]
(e) Repeat each set of readings and calculate an average to reduce the effect of random errors. Other valid answers: use a digital ammeter for more precise readings, or use a variable resistor to keep the current low and avoid heating effects. [1]
(f) The ammeter readings (0.14, 0.07, 0.05, 0.04 A) show values to the nearest 0.01 A. The resolution is 0.01 A. [1]
(g) The voltmeter must be connected in parallel with (across) the resistors being tested, so that it measures the potential difference across them. [1]
(h) Contact resistance at the connections between resistors adds extra, unmeasured resistance. Other valid answers: difficulty reading the ammeter accurately at low currents, or the internal resistance of the battery affecting the terminal voltage. [1]
Why this matters: In a series circuit, the total resistance equals the sum of the individual resistances: \( R_{\text{total}} = R_1 + R_2 + R_3 + \ldots \). Each resistor opposes current flow, and connecting more in series increases the total opposition. The current drops because the same driving voltage must push charge through an ever-increasing resistance. The voltmeter reading barely changes because the battery's e.m.f. is nearly constant, but the current falls as resistance grows. This principle is fundamental to understanding series circuits in everyday applications such as Christmas tree lights wired in series.
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