A student investigates resistors connected in series in an electrical circuit - Edexcel - GCSE Physics - Question 6 - 2020 - Paper 1
Question 6
A student investigates resistors connected in series in an electrical circuit.
The student has
- a 3.0V battery
- a 22Ω resistor
- a resistor marked X.
The student... show full transcript
Worked Solution & Example Answer:A student investigates resistors connected in series in an electrical circuit - Edexcel - GCSE Physics - Question 6 - 2020 - Paper 1
Step 1
Describe how the student should correct the mistake.
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Answer
The student should move the voltmeter to be in parallel with the resistor marked X. This will allow for the accurate measurement of the voltage across resistor X.
Step 2
State the value of the voltage across the 22Ω resistor.
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Answer
The voltage across the 22Ω resistor is 0.9 V.
Step 3
The current in resistor X is 0.041 A. Show that the resistance of resistor X must be about 50 ohms. Use the equation V = I × R.
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Answer
Using the given voltage across resistor X, 2.1 V, and the current, 0.041 A, we apply the formula:
R=IV=0.041 A2.1 V≈51.2Ω
This rounds approximately to 50 ohms.
Step 4
Calculate the power in resistor X when the voltage across X is 2.1V and the current in resistor X is 0.041 A.
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The power in resistor X can be calculated using the formula:
P=V×I=2.1 V×0.041 A≈0.086 W
Thus, the power is approximately 0.086 W.
Step 5
Calculate the overall resistance of the 22 ohm resistor and resistor X.
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Answer
The total resistance in a series circuit is the sum of individual resistances:
Rtotal=R22Ω+RX=22Ω+51.2Ω≈73.2Ω
Step 6
The current in the circuit is 0.041 A. Calculate the energy transferred in 2 minutes. Use the equation E = I × V × t.
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Answer
To calculate the energy transferred, we convert 2 minutes to seconds (2 minutes = 120 seconds). Thus, we use the formula:
