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A resistor is connected to a power supply - Edexcel - GCSE Physics - Question 8 - 2018 - Paper 1

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A resistor is connected to a power supply. The potential difference across the resistor is 6.0V. (i) Which of these corresponds to a potential difference of 6.0V? ... show full transcript

Worked Solution & Example Answer:A resistor is connected to a power supply - Edexcel - GCSE Physics - Question 8 - 2018 - Paper 1

Step 1

Which of these corresponds to a potential difference of 6.0V?

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Answer

The correct answer is C 6.0 joules per coulomb. This is because the unit of potential difference (voltage) can also be expressed as joules per coulomb.

Step 2

Calculate, in minutes, the time taken for this amount of charge to flow through the resistor.

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Answer

To find the time taken for a charge of 42C to flow at a current of 200 mA, we use the formula:

t = rac{Q}{I}

Substituting in the values:

t = rac{42C}{0.2A} = 210 seconds

To convert seconds to minutes:

t = rac{210}{60} = 3.5 ext{ minutes}

Step 3

Calculate the total energy transferred by the 6.0V power supply when a charge of 42C flows through the resistor.

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Answer

The energy transferred can be calculated using the formula:

E=VimesQE = V imes Q

Substituting in the values:

E=6.0Vimes42C=252extjoulesE = 6.0V imes 42C = 252 ext{ joules}

Step 4

Explain why the resistor becomes warm.

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Answer

The resistor becomes warm due to the collisions between moving electrons and the lattice structure of the resistor. As electrons flow through the resistor, they collide with atoms in the lattice. These collisions transfer energy to the atoms, causing them to vibrate more, which in turn increases the thermal energy of the resistor, resulting in a warmer temperature.

Step 5

Deduce how the resistors have been arranged inside the cardboard tube.

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Answer

To find the arrangement of the resistors, we first need to calculate the total resistance.

Given:

  • Potential difference (V) = 6.0V
  • Current (I) = 1.2A

Using Ohm's law:

R = rac{V}{I} = rac{6.0V}{1.2A} = 5 ext{ ohms}

Each resistor has a resistance of 100 ohms. If they were in series, the total resistance would be:

Rtotalext(series)=100+100=200extohmsR_{total ext{ (series)}} = 100 + 100 = 200 ext{ ohms}

Since 5 ohms is less than 200 ohms, the resistors must be connected in parallel. This arrangement allows for a lower total resistance, thus supporting the current of 1.2A.

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