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
a) Describe how the student should correct the mistake.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The student should move the voltmeter so that it is connected in parallel with the resistor marked X. This will allow for the correct measurement of the voltage across the resistor.
Step 2
b) i) State the value of the voltage across the 22Ω resistor.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The voltage across the 22Ω resistor is 0.9V.
Step 3
b) ii) Show that the resistance of resistor X must be about 50 ohms.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
We are given the voltage across resistor X, V = 2.1V, and the current through resistor X, I = 0.041A.
Using the formula:
V=IimesR
We can rearrange to find R:
R=IV=0.0412.1≈51.22Ω
This value rounds to approximately 50Ω.
Step 4
b) iii) Calculate the power in resistor X.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The power in resistor X can be calculated using the formula:
P=VimesI
Substituting the values:
P=2.1Vimes0.041A=0.0861W
Thus, the power is approximately 0.086W.
Step 5
b) iv) Calculate the overall resistance of the 22 ohm resistor and resistor X.
97%
117 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The overall resistance in a series circuit is the sum of the individual resistances. Thus:
Rtotal=R22+RX=22+51.22approx73.22Ω
Step 6
b) v) Calculate the energy transferred in 2 minutes.
97%
121 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Using the equation for energy:
E=I×V×t
Where I = 0.041A, V = 3.0V, and t = 2 imes 60s = 120s,
We substitute:
E=0.041×3.0×120=14.76J
Thus, the energy transferred is approximately 15J.
