Two resistors X and Y are connected in series with a power supply of emf 30 V and negligible internal resistance - AQA - A-Level Physics - Question 29 - 2021 - Paper 1
Question 29
Two resistors X and Y are connected in series with a power supply of emf 30 V and negligible internal resistance.
The resistors are made from wire of the same materi... show full transcript
Worked Solution & Example Answer:Two resistors X and Y are connected in series with a power supply of emf 30 V and negligible internal resistance - AQA - A-Level Physics - Question 29 - 2021 - Paper 1
Step 1
Calculate the Resistance of Resistor X
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 resistance of a wire is given by the formula:
R = rac{ρL}{A}
where ρ is the resistivity, L is the length, and A is the cross-sectional area of the wire. For Resistor X, using diameter d:
A_X = rac{π(d/2)^2}{4} = rac{πd^2}{4}
Thus, the resistance of Resistor X becomes:
R_X = rac{ρL}{(πd^2 / 4)} = rac{4ρL}{πd^2}
Step 2
Calculate the Resistance of Resistor Y
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
For Resistor Y, using diameter 2d:
A_Y = rac{π(2d/2)^2}{4} = rac{π(2d)^2}{4} = πd^2
Thus, the resistance of Resistor Y becomes:
R_Y = rac{ρL}{(πd^2)} = rac{ρL}{πd^2}
Step 3
Determine the Total Resistance in the Circuit
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
Since X and Y are in series:
R_{total} = R_X + R_Y = rac{4ρL}{πd^2} + rac{ρL}{πd^2} = rac{5ρL}{πd^2}
Step 4
Calculate the Current in the Circuit
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
Using Ohm's law, the total current I from the power supply of 30 V is:
I = rac{V}{R_{total}} = rac{30 V}{ rac{5ρL}{πd^2}} = rac{30πd^2}{5ρL} = rac{6πd^2}{ρL}
Step 5
Calculate the Voltage Drop Across 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 voltage drop V_X across Resistor X can be calculated as:
V_X = I imes R_X = rac{6πd^2}{ρL} imes rac{4ρL}{πd^2} = 24 V
Step 6
Conclusion
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
Thus, the voltage reading on the voltmeter, which is across Resistor Y, is:
VY=V−VX=30V−24V=6V
Therefore, the reading on the voltmeter is 6 V.
