MCQ on electricity

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fluffypoopies
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#1
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#1
Why is the answer to this question C?
Name:  Screenshot 2022-05-21 at 20-16-28 Question paper (A-level) Paper 1 - November 2020 - AQA-74081-Q.png
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Could someone give me advice on how I should go about approaching this question because I'm stumped lol
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Driving_Mad
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#2
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Lost volts = 2V/3. Current = pd across 2 ohm resistor / 2ohms, so V/3/2. Does that help you to work out the internal resistance using the formulae?
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fluffypoopies
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(Original post by Driving_Mad)
Lost volts = 2V/3. Current = pd across 2 ohm resistor / 2ohms, so V/3/2. Does that help you to work out the internal resistance using the formulae?
ohhh thanks a lot. Why is the lost volts 2v/3 and not v/3 if the voltmeter is applied across the internal resistance? - I kept using lost volts as v/3 which was where I kept messing up.
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Driving_Mad
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(Original post by fluffypoopies)
ohhh thanks a lot. Why is the lost volts 2v/3 and not v/3 if the voltmeter is applied across the internal resistance? - I kept using lost volts as v/3 which was where I kept messing up.
The voltmeter records the terminal pd. You can never get a direct reading off the voltmeter of the lost volts.

If you connected the voltmeter across the resistor, you’d still get V/3 as that’s the terminal pd.
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username5318474
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(Original post by fluffypoopies)
ohhh thanks a lot. Why is the lost volts 2v/3 and not v/3 if the voltmeter is applied across the internal resistance? - I kept using lost volts as v/3 which was where I kept messing up.
When a voltmeter is connected across the battery, it always measures the terminal pd.

Instead of working out the current, you can use the potential divider argument to determine r:
If 2 Ohms recieves V/3
2V/3 must be given to an internal reistor with 4 Ohms resistance.

But you can go about this by determining the current in the circuit and applying V = Ir. This is just a simpler alternative that consoldiates knowledge on potential dividers.
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fluffypoopies
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(Original post by speed_bird)
When a voltmeter is connected across the battery, it always measures the terminal pd.

Instead of working out the current, you can use the potential divider argument to determine r:
If 2 Ohms recieves V/3
2V/3 must be given to an internal reistor with 4 Ohms resistance.

But you can go about this by determining the current in the circuit and applying V = Ir. This is just a simpler alternative that consoldiates knowledge on potential dividers.
thank you for this
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