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Take the ammeter and voltmeter readings as shown in the circuit diagram. Record your results in the table below. Write a conclusion for your experiment. Refer to the potential difference across resistors in parallel and the sum of the current through resistors in parallel.

Physics

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Answer

3.5

0.3 A\text{0.3 A}

Step-by-step solution

3 steps

Part 3

Part 3.1

One precaution that needs to be taken when carrying out an experiment is to ensure all electrical connections are tight and secure to prevent inaccurate readings or short circuits.

Part 3.2

  • 3.2.1 A switch: A switch is used to open or close an electrical circuit, thereby controlling the flow of current.
  • 3.2.2 An ammeter: An ammeter is used to measure the electric current flowing through a specific point in a circuit. It must be connected in series.
  • 3.2.3 A battery: A battery provides the electromotive force (EMF) or potential difference needed to drive electric current through a circuit.

Part 3.3

Electric current is the rate of flow of electric charge through a conductor.

Part 3.4

One coulomb per second is defined as an Ampere.

Part 3.5

0.3 A\text{0.3 A}

  1. Convert the time from minutes to seconds.

    t=20 minutes×60secondsminute=1200 st = 20 \text{ minutes} \times 60 \frac{\text{seconds}}{\text{minute}} = 1200 \text{ s}
  2. Use the formula for current, I=QtI = \frac{Q}{t}, where QQ is charge and tt is time.

    I=360 C1200 sI = \frac{360 \text{ C}}{1200 \text{ s}}
  3. Calculate the current.

    I=0.3 AI = 0.3 \text{ A}

    The current flowing through the bulb is 0.3 A\boxed{\text{0.3 A}}.

Part 3.6

The question is incomplete as it describes an experiment and provides data but does not state what needs to be done with the data or what question needs to be answered.

However, based on the provided data for the nichrome wire:

  • Current (A): 0.2, 0.4, 0.6, 0.8, 1.0
  • Potential difference (V): 0.8, 1.6, 2.4, 3.2, 4.0

We can observe that for the nichrome wire, as the current increases, the potential difference across it also increases proportionally. If we calculate the ratio of potential difference to current (V/IV/I), we get:

  • 0.8 V/0.2 A=4Ω0.8 \text{ V} / 0.2 \text{ A} = 4 \Omega
  • 1.6 V/0.4 A=4Ω1.6 \text{ V} / 0.4 \text{ A} = 4 \Omega
  • 2.4 V/0.6 A=4Ω2.4 \text{ V} / 0.6 \text{ A} = 4 \Omega
  • 3.2 V/0.8 A=4Ω3.2 \text{ V} / 0.8 \text{ A} = 4 \Omega
  • 4.0 V/1.0 A=4Ω4.0 \text{ V} / 1.0 \text{ A} = 4 \Omega

This shows that the resistance of the nichrome wire is constant at 4Ω4 \Omega, which is consistent with Ohm's Law for an ohmic conductor at constant temperature. To compare the effect of the type of conductor (nichrome vs. copper) on current strength, we would need the potential difference readings for the copper wire at the same current values.

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