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Electricity Multiple choice question from OCR
Page 7 question 9 here https://www.ocr.org.uk/Images/409017-question-paper-unit-h156-01-breadth-in-physics.pdf
About the 9V battery connected to two resistors...
I've tried a few different methods to solve it and the closest I can get to the right answer is 6.75 mW using P=IV (and the fact that V will be 3/4 of the input voltage of 9V). The correct answer is A but I'm assuming since my answer rounds to 7 my method is not correct. Anyone able to help?
About the 9V battery connected to two resistors...
I've tried a few different methods to solve it and the closest I can get to the right answer is 6.75 mW using P=IV (and the fact that V will be 3/4 of the input voltage of 9V). The correct answer is A but I'm assuming since my answer rounds to 7 my method is not correct. Anyone able to help?
If you calculate the power in the two resistors individually, using I2R and add them...
(current in upper resistor is 3mA and lower one is 2mA (from 3mA -1mA to the other circuit)
Then calculate the power delivered by the cell from P=VI (I = 3mA)
The difference is the power delivered to the other circuit.
(current in upper resistor is 3mA and lower one is 2mA (from 3mA -1mA to the other circuit)
Then calculate the power delivered by the cell from P=VI (I = 3mA)
The difference is the power delivered to the other circuit.
Original post by DRDANDY
Page 7 question 9 here https://www.ocr.org.uk/Images/409017-question-paper-unit-h156-01-breadth-in-physics.pdf
About the 9V battery connected to two resistors...
I've tried a few different methods to solve it and the closest I can get to the right answer is 6.75 mW using P=IV (and the fact that V will be 3/4 of the input voltage of 9V). The correct answer is A but I'm assuming since my answer rounds to 7 my method is not correct. Anyone able to help?
About the 9V battery connected to two resistors...
I've tried a few different methods to solve it and the closest I can get to the right answer is 6.75 mW using P=IV (and the fact that V will be 3/4 of the input voltage of 9V). The correct answer is A but I'm assuming since my answer rounds to 7 my method is not correct. Anyone able to help?
I'm reviving this post because I don't understand one thing about it....
How can these two statements yield different answers?;
'the ratio of resistances means the 3000 ohm resistor is getting 3/4 of the voltage which = 6.75 V'
vs
'Using V=IR, the remaining current is 2mA and the Resistance is 3000 ohms so then the voltage = 6.00V'
How can they both be true, or which one is wrong?
Original post by DRDANDY
I'm reviving this post because I don't understand one thing about it....
How can these two statements yield different answers?;
'the ratio of resistances means the 3000 ohm resistor is getting 3/4 of the voltage which = 6.75 V'
vs
'Using V=IR, the remaining current is 2mA and the Resistance is 3000 ohms so then the voltage = 6.00V'
How can they both be true, or which one is wrong?
How can these two statements yield different answers?;
'the ratio of resistances means the 3000 ohm resistor is getting 3/4 of the voltage which = 6.75 V'
vs
'Using V=IR, the remaining current is 2mA and the Resistance is 3000 ohms so then the voltage = 6.00V'
How can they both be true, or which one is wrong?
The first one is wrong.
Current in upper resistor is 3mA and V=IR gives a pd of 3V across it
Current in the lower resistor is 2mA because 1mA goes into the circuit with X and Y
So pd across the lower resistor is 6V using V=IR for it (and because it must have the rest of the 9V from the battery)
The reason the first statement was wrong, is that the ratio rule used would only be true if the two resistors were all that was in the circuit and both resistors had the same current. But they don't, because some current goes to X.
use p=i^2(r) to solve for the power between x and y.
so current through x-y is 2mA as 1mA is drawn out the circuit at x, resistance through x-y is just 3,000ohms
Using the equation you can convert mA to amps with x10-3, you get 0.012W and in mW 12mW
so current through x-y is 2mA as 1mA is drawn out the circuit at x, resistance through x-y is just 3,000ohms
Using the equation you can convert mA to amps with x10-3, you get 0.012W and in mW 12mW
(edited 10 months ago)
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