# 3 Hard Physics electricity questions

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Hey... help needed if anyone can answer these, i think there some of the harder PHYA1 electricity questions.

Firstly, on the Jan 2011 PHYA1 paper,

(circuit resistance decreases)

(supply) current increases or potential divider argument, hence pd across 540 Ω resistor increases

hence pd across 1200 Ω decreases

or resistance in parallel combination decreases

pd across parallel resistors decreases

And finally (i promise) from the Nelson Thornes AQA Unit 1 practise paper thing, Question 17 shows a circuit, with internal resistance r, connected to an oscilloscope, and when the circuit is connected to a resistor of resistance 15Ohms, the pd falls.

Why??????? the answers says it's because current flows through the resistor and pd is lost across the internal resistance of the battery? but this makes no sense to me if resistance has increased, by V = IR shouldn't PD also increase when current flows through the resistor?

Firstly, on the Jan 2011 PHYA1 paper,

**Assuming a lamp is connected to a 12V supply with negligible internal resistance, state and explain the changes, if any, to the final current through the lamp if it is connected to the same supply with another lamp i) in series, ii) in parallel**

__Initially i thought, as we've always been taught in this module, that the current with the lamps in series will be equal and when parallel the current is shared? (Kirchoffs laws?) but the mark scheme says completely the opposite :s__

Secondly, again on the Jan 2011 PHYA1 paper, (see attached), the very last question asks:*The temperature of the thermistor is increased so that resistance decreases. State and explain what happens to the pd across the 1200Ohm resistor?*

__The answer says:__

(voltage of supply constant)(circuit resistance decreases)

(supply) current increases or potential divider argument, hence pd across 540 Ω resistor increases

hence pd across 1200 Ω decreases

or resistance in parallel combination decreases

pd across parallel resistors decreases

__I don't get why the pd across the 1200Ohm resistor decreases?? I understand that the Pd across the 540Ohm resistor increases, but why is it different for the 1200Ohm resistor??__

And finally (i promise) from the Nelson Thornes AQA Unit 1 practise paper thing, Question 17 shows a circuit, with internal resistance r, connected to an oscilloscope, and when the circuit is connected to a resistor of resistance 15Ohms, the pd falls.

Why??????? the answers says it's because current flows through the resistor and pd is lost across the internal resistance of the battery? but this makes no sense to me if resistance has increased, by V = IR shouldn't PD also increase when current flows through the resistor?

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#2

First one is asking about the change in the current

you seem to be answering a different question about the difference between the current in the two bulbs or the current in other parts of the circuit.

in the series arrangement the current

in the parallel arrangement the current

you aren't asked about the current in the rest of the circuit or the difference between the currents in the bulbs - obv try to answer the question you are actually asked

**in one lamp**when you connect another identical lamp in series using the same supply.you seem to be answering a different question about the difference between the current in the two bulbs or the current in other parts of the circuit.

in the series arrangement the current

**in one lamp**goes down compared to one lamp on its own because there is now more resistance in the circuit and the pd across it's terminals is reduced.in the parallel arrangement the current

**in one lamp**stays the same because the pd across it's terminals stays the same.you aren't asked about the current in the rest of the circuit or the difference between the currents in the bulbs - obv try to answer the question you are actually asked

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#3

For the first question you have to take into account that adding another lamp to the circuit will change the total resistance. If you add another lamp in series, the total resistance will double so current in the circuit will be halved. But if you add another lamp in parallel the total resistance of the circuit will be halved (you can put values in the equation and check), so the current will be doubled. But then as its a parallel circuit, the current will split up in half in each branch again so basically there will be no change in the current through the lamp.

For the second question, I use the potential divider argument. If the resistance of the thermistor is decreased, the resistance of the entire parallel combination will decrease. The parallel combination is in series with the 540 ohm resistor so the parallel combination will now have a smaller share of the total pd. Pd is the same in both branches of the parallel bit so pd across the 1200 ohm resistor will decrease.

The third one is your basic V= Emf - Ir situation. When S2 is open there is no current in the main circuit, so V= Emf. When S2 is closed, there now is a current in the main circuit so a pd will be induced across the internal resistance and V= Emf - Ir. In terms of resistance - if you think about it, the voltmeter has infinite resistance. If you then add any other component in parallel (by closing S2) the total resistance will decrease greatly and so a current can now flow.

Hope that helped a little

For the second question, I use the potential divider argument. If the resistance of the thermistor is decreased, the resistance of the entire parallel combination will decrease. The parallel combination is in series with the 540 ohm resistor so the parallel combination will now have a smaller share of the total pd. Pd is the same in both branches of the parallel bit so pd across the 1200 ohm resistor will decrease.

The third one is your basic V= Emf - Ir situation. When S2 is open there is no current in the main circuit, so V= Emf. When S2 is closed, there now is a current in the main circuit so a pd will be induced across the internal resistance and V= Emf - Ir. In terms of resistance - if you think about it, the voltmeter has infinite resistance. If you then add any other component in parallel (by closing S2) the total resistance will decrease greatly and so a current can now flow.

Hope that helped a little

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(Original post by

For the second question, I use the potential divider argument. If the resistance of the thermistor is decreased, the resistance of the entire parallel combination will decrease. The parallel combination is in series with the 540 ohm resistor so the parallel combination will now have a smaller share of the total pd. Pd is the same in both branches of the parallel bit so pd across the 1200 ohm resistor will decrease.

Hope that helped a little

**Aspire101**)For the second question, I use the potential divider argument. If the resistance of the thermistor is decreased, the resistance of the entire parallel combination will decrease. The parallel combination is in series with the 540 ohm resistor so the parallel combination will now have a smaller share of the total pd. Pd is the same in both branches of the parallel bit so pd across the 1200 ohm resistor will decrease.

Hope that helped a little

I understand that the entire parallel resistance combination will decrease... does that mean that as V = IR, the the pd of the entire parallel combination also decreases, thus as V from the battery stays the same, the PD across the 540Ohm resistor has to increase to compensate?

As you can see i don't have a strong grasp of the potential divider argument!

I'm having a very similar problem with a question in jan 2012 paper... mind if i PM you very quickly?

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#5

__1st one__

**Series**Two lamps connected in series will have a greater combined resistance. Since, the resistance is greater, lesser current can flow through.

**Parallel**For two lamps connected in parallel voltage is the same for both. Since the lamps are identical, they should have same resistance. So they have same current.

I= V1/R1 = v2/R2

__3rd on__I have no clue what oscilliscope is(measures oscillations?). I do WJEC sorry!

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#6

(Original post by

That helps a lot thanks, i get the last one, still don't understand the lamp one but i'm just going to forget that - it's the potential divider one i'm seriously stuck on.

I understand that the entire parallel resistance combination will decrease... does that mean that as V = IR, the the pd of the entire parallel combination also decreases, thus as V from the battery stays the same, the PD across the 540Ohm resistor has to increase to compensate?

As you can see i don't have a strong grasp of the potential divider argument!

I'm having a very similar problem with a question in jan 2012 paper... mind if i PM you very quickly?

**physicso**)That helps a lot thanks, i get the last one, still don't understand the lamp one but i'm just going to forget that - it's the potential divider one i'm seriously stuck on.

I understand that the entire parallel resistance combination will decrease... does that mean that as V = IR, the the pd of the entire parallel combination also decreases, thus as V from the battery stays the same, the PD across the 540Ohm resistor has to increase to compensate?

As you can see i don't have a strong grasp of the potential divider argument!

I'm having a very similar problem with a question in jan 2012 paper... mind if i PM you very quickly?

Yes exactly - total pd is constant so if the pd across the parallel section decreases, pd across the 540 ohm resistor must increase to compensate

And sure you can PM me you like - no problem

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