GCSE PHY Lamp Current, potential difference and resistance

A student investigated how the current in a filament lamp varies with the potential difference across the filament lamp. What does the graph show about the relationship between current and potential difference?

• The current increases linearly with potential difference.

• The current decreases linearly with potential difference.

• The current increases non-linearly with potential difference.

• The current remains constant with potential difference.

Which two statements form part of the METHOD to investigate a lamp characteristics ?

•  ammeter in series with filament lamp

•  ammeter in parallel with filament lamp

•  current measured with an ammeter

• No relevant content current measured with an voltmeter

which statement is NOT a valid step in the METHOD to investigate a lamp filament characteristics ?

• discard anomalies

•  variable resistor (or variable power pack or variable number of cells) used to vary current in and p.d. across filament lamp

• take repeat readings and calculate a mean

• read the voltage p.d, and on the variable resistor

What is measured with an ammeter in the investigation?

• Voltage

• Current

• Resistance

• Power

Which is the correct circuit that should be used in the investigation?

• A

• B

• C

What is the range of potential difference (p.d.) mentioned in the lamp filament investigation?

• 0 to 3 V

• 0 to 6 V

• 0 to 9 V

• 0 to 12 V

What is the interval of potential difference (p.d.) in the readings of voltage from point to point?

• 0.5 V

• 1 V

• 1.5 V

• 2 V

What should be done to obtain negative values according to the indicative content?

• Increase the current

• Decrease the voltage

• Reverse connections to power supply

• Use a different ammeter

What is the correct value of resistance (R) if 3.0 = 0.16 × R?

• 18.75 Ω

• 19 Ω

• 18.8 Ω

• 20 Ω

Find the two workings that are correct to find the energy transferred.

• Q= I t

  E= QV

  Q = 0.21 × 1800

  E = 378 × 6.0 joules

• Q= I t

  E= I V

  Q = 0.21 × 1800

  E = 378 × 6.0 joules

• P= VI

  E= P t

  P = 0.21 × 6.0

  E = 1.26 × 1800 joules

• P= VI

  E= P /t

  P = 0.21 × 6.0

  E = 1.26 × 1800 joules