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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 B 6.0 amps per coulomb, as it is derived from the definition of potential difference, which states that 1 volt is equal to 1 joule 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

First, convert the current from mA to A:

200extmA=0.2extA200 ext{ mA} = 0.2 ext{ A}

Using the formula for current, where current (I) is charge (Q) divided by time (t):

I=Qtt=QII = \frac{Q}{t} \\ t = \frac{Q}{I}

Substituting the values, we have:

t=42extC0.2extA=210extsecondst = \frac{42 ext{ C}}{0.2 ext{ A}} = 210 ext{ seconds}

To convert seconds to minutes:

t=21060=3.5extminutest = \frac{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 (E) can be calculated using the formula:

E=V×QE = V × Q

where V is the potential difference and Q is the charge. Substituting the values:

E=6.0extV×42extC=252extJE = 6.0 ext{ V} × 42 ext{ C} = 252 ext{ J}

Step 4

Explain why the resistor becomes warm.

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Answer

The resistor becomes warm due to the collisions between the flowing electrons and the atoms in the resistive material. As the electrons move through the lattice, they collide with the vibrating atoms, which transfers energy and increases the temperature of the resistor.

Step 5

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

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Answer

Given the potential difference of 6.0V and the current of 1.2A, we can find the equivalent resistance (R) between points P and Q using Ohm's law:

R = \frac{V}{I} = \frac{6.0 ext{ V}}{1.2 ext{ A}} = 5 ext{ ohms} $$ If the resistors were in series, their total resistance would be the sum: $$ 100 ext{ ohms} + 100 ext{ ohms} = 200 ext{ ohms} \\ \text{(current would be 0.3 A).} $$ However, 0.3 A is less than 1.2 A, indicating that the total potential difference is less than 12 V. Hence, the resistors must be connected in parallel to allow the higher current.

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