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A student investigates resistors connected in parallel using a number of resistors - Edexcel - GCSE Physics Combined Science - Question 5 - 2020 - Paper 1

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A student investigates resistors connected in parallel using a number of resistors. Each resistor has the same resistance. Figure 10 shows a circuit diagram with on... show full transcript

Worked Solution & Example Answer:A student investigates resistors connected in parallel using a number of resistors - Edexcel - GCSE Physics Combined Science - Question 5 - 2020 - Paper 1

Step 1

Add to Figure 10:

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Answer

  1. Draw a voltmeter connected in parallel across resistor R to measure the potential difference.
  2. Add a second resistor in parallel with resistor R.

Step 2

State the measurements that the student must take to find the overall resistance of the resistors in parallel.

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Answer

  1. Measure the potential difference (voltage) across the resistors in parallel.
  2. Measure the total current flowing through the circuit.

Step 3

State the resistance of a single resistor.

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Answer

The resistance of a single resistor is 1 Ω.

Step 4

Comment on the relationship between the overall resistance of the circuit and the number of resistors in parallel.

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The overall resistance decreases as the number of resistors in parallel increases. This decrease is non-linear, meaning that although resistance continues to drop with additional resistors, the rate of decrease slows down. This indicates an inverse proportionality between the overall resistance and the number of parallel resistors, supported by the graph in Figure 11.

Step 5

Calculate the potential difference across the 15Ω resistor.

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Answer

Using Ohm's Law, the potential difference (V) can be calculated as:

V=I×RV = I \times R Here, I = 0.20 A and R = 15 Ω, thus, V=0.20×15=3 VV = 0.20 \times 15 = 3 \text{ V}

Step 6

Calculate the total power dissipated when there is a current of 0.20 A in the two resistors.

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Answer

Using the formula for power:

P=I×RP = I \times R The total resistance (R) is the sum of both resistors, 15 Ω + 20 Ω = 35 Ω. Thus:

P=0.202×35=1.4 WP = 0.20^2 \times 35 = 1.4 \text{ W}

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