In this experiment, you will investigate the charge delivered by a battery. You connects a 22 Ω resistor to a 4.5 V battery, a switch and an ammeter. You cl...
Question 1 Report
In this experiment, you will investigate the charge delivered by a battery. You connects a 22 Ω resistor to a 4.5 V battery, a switch and an ammeter. You closes the switch and records the ammeter reading. You then calculates the charge delivered in three different time intervals using the equation Q = I × t. The ammeter reads 0.20 A throughout. Your results are shown in Table 10.1.
| Time t / s | Current I / A | Charge Q / C |
|---|---|---|
| 30 | 0.20 | |
| 60 | 0.20 | |
| 120 | 0.20 |
You uses a digital ammeter with a resolution of 0.01 A for accuracy. Before starting the timing, you checks the ammeter reads zero with the switch open, then closes the switch and starts the stopwatch simultaneously. You keeps the switch closed throughout the experiment. You records the ammeter reading periodically to verify it stays constant. The battery voltage is 4.5 V, which produces a current of 0.20 A through the 22 ohm resistor.
(a) Complete Table 10.1 by calculating the charge Q for each time. [2]
(b) Record the ammeter reading. [1]
(c) Describe the relationship between time and charge delivered. [1]
(d) Calculate the charge that would be delivered in 5 minutes. [1]
(e) Measure the total charge delivered in the first 120 seconds. [1]
(f) State one assumption made in the calculation. [1]
Answer Details
(a) Using \( Q = I \times t \) with \( I = 0.20 \) A: [2]
| Time \( t \) / s | Current \( I \) / A | Charge \( Q \) / C |
|---|---|---|
| 30 | 0.20 | \( 0.20 \times 30 = 6.0 \) |
| 60 | 0.20 | \( 0.20 \times 60 = 12 \) |
| 120 | 0.20 | \( 0.20 \times 120 = 24 \) |
All three values calculated correctly. [1] Correct method: multiply current by time. [1]
(b) The ammeter reading is 0.20 A. [1]
(c) Charge is directly proportional to time. As time doubles, charge doubles (e.g. 6.0 C at 30 s, 12 C at 60 s, 24 C at 120 s). [1]
This is expected from \( Q = It \) with constant \( I \): Q increases linearly with t.
(d) Charge delivered in 5 minutes:
\[ t = 5 \times 60 = 300 \text{ s} \]
\[ Q = 0.20 \times 300 = 60 \text{ C} \] [1]
(e) The total charge delivered in the first 120 seconds is 24 C. [1]
(f) The calculation assumes that the current remains constant throughout the experiment (i.e. the battery voltage does not drop and the resistance does not change). [1]
Download The App On Google Playstore
Everything you need to excel in your exams
