2395 Question hope someone can help

[Gtbadboy1]

Determine the actual voltage drop, in volt, if the combined resistance of the line and neutral conductors, at 20 degrees C is 0.45 ohms and the load current is 20A.

 

[Deleted member 9648]

Ohms law.

 

[Simonslimline]

VD = (R1+RN) x Ib X 1.2

 

[davesparks]

You need the supply voltage to be able to answer this

 

[davesparks]

VD = (R1+RN) x Ib X 1.2

What is the 1.2?

 

[Simonslimline]

What is the 1.2?

The multiplier to compensate for the difference in conductor temperature at the time of the test and the normal operating temperature of the conductors when maximum resistance and max volt drop will occur.

Thats how its written in the exam success book for the 2394/5

20c + 50 x 0.004 = 70c.

 

[telectrix]

You need the supply voltage to be able to answer this

why?//

 

[telectrix]

The multiplier to compensate for the difference in conductor temperature at the time of the test and the normal operating temperature of the conductors when maximum resistance and max volt drop will occur.

Thats how its written in the exam success book for the 2394/5

20c + 50 x 0.004 = 70c.

and how do you know the operating temperature. it could be 25mm cable @ 20A. never get warm, even.

 

[Simonslimline]

and how do you know the operating temperature. it could be 25mm cable @ 20A. never get warm, even.

It's what they are looking for in the exam.

It is one of the questions that pops up.

 

[davesparks]

why?//

Cos I didn't read the question properly

 

[davesparks]

The multiplier to compensate for the difference in conductor temperature at the time of the test and the normal operating temperature of the conductors when maximum resistance and max volt drop will occur.

Thats how its written in the exam success book for the 2394/5

20c + 50 x 0.004 = 70c.

So what is the 0.004?

 

[Simonslimline]

So what is the 0.004?

The simplified resistance coefficient per °C at 20°C given by BS 6360 for copper and aluminium conductors.

 

[marconi]

Determine the actual voltage drop, in volt, if the combined resistance of the line and neutral conductors, at 20 degrees C is 0.45 ohms and the load current is 20A.

By Marconi Here are some thoughts on how to read the question.

Determine is a long word for calculate or use a formula.

The answer is a number in the units of Volts.

'the combined resistance of the line and neutral conductors' - this tells you that there are two conductors supplying the load (and it is single phase) and if the the line has a resistance of Rline and the neutral has a resistance of Rneut then Rline + Rneut = 0.45 Ohms.

Why have they mentioned 20 degrees C? This is because the resistance of a conductor depends on it is temperature. For metallic conductors the higher the temperature the higher the resistance and the lower the temperature the lower the resistance. So to tell you something truthful and in fullest detail they give the resistance for the actual temperature of the conductors. Mentioning the temperature with a figure is a way the examiner is testing your understanding of the resistance of conductors and giving a number which you may or may not need to use in the calculation on actual voltage drop. It is the temperature of the conductors while 20Amps is flowing through them both.

Whether the voltage drop takes place equally between line and neutral conductors does not matter; it is the total effect of resistance between the supply and the load which determines the voltage drop.

So, Supply - resistance of line and neutral conductors - load.

Then you are told the current which is drawn by the load which is 20A. This current flows through the load and also through the resistances of the line and neutral conductors. We know that what I have underlined is 0.45 Ohms.

The 'Volt drop' they are asking about is across the resistance of the line and neutral conductors.

So, in your mind , and in a diagram if you drew one - which you should always - you have a resistance of 0.45 Ohm and through it is passing a current of 20 A. How do you calculate the voltage drop across the resistor? Remembering Ohm's Law you know V=I x R.

So Vdrop is 20 x 0.45= 9Volts. Don't forget to put in Volts or V after the number if you wamt full marks.

You do not need to use the temperature in your calculations.

 

[Richard Burns]

Good clear answer above.
From the other posts and the fact that the question asks for actual volt drop, I would assume they mean under maximum operating conditions and so I think you would need to take the temperature into consideration.

 

[Deleted member 9648]

By Marconi Here are some thoughts on how to read the question.

Determine is a long word for calculate or use a formula.

The answer is a number in the units of Volts.

'the combined resistance of the line and neutral conductors' - this tells you that there are two conductors supplying the load (and it is single phase) and if the the line has a resistance of Rline and the neutral has a resistance of Rneut then Rline + Rneut = 0.45 Ohms.

Why have they mentioned 20 degrees C? This is because the resistance of a conductor depends on it is temperature. For metallic conductors the higher the temperature the higher the resistance and the lower the temperature the lower the resistance. So to tell you something truthful and in fullest detail they give the resistance for the actual temperature of the conductors. Mentioning the temperature with a figure is a way the examiner is testing your understanding of the resistance of conductors and giving a number which you may or may not need to use in the calculation on actual voltage drop. It is the temperature of the conductors while 20Amps is flowing through them both.

Whether the voltage drop takes place equally between line and neutral conductors does not matter; it is the total effect of resistance between the supply and the load which determines the voltage drop.

So, Supply - resistance of line and neutral conductors - load.

Then you are told the current which is drawn by the load which is 20A. This current flows through the load and also through the resistances of the line and neutral conductors. We know that what I have underlined is 0.45 Ohms.

The 'Volt drop' they are asking about is across the resistance of the line and neutral conductors.

So, in your mind , and in a diagram if you drew one - which you should always - you have a resistance of 0.45 Ohm and through it is passing a current of 20 A. How do you calculate the voltage drop across the resistor? Remembering Ohm's Law you know V=I x R.

So Vdrop is 20 x 0.45= 9Volts. Don't forget to put in Volts or V after the number if you wamt full marks.

You do not need to use the temperature in your calculations.

An extremely long winded but far better explanation of what I already stated in post #2!

 

[marconi]

By Marconi Here are some thoughts on how to read the question.

Determine is a long word for calculate or use a formula.

The answer is a number in the units of Volts.

'the combined resistance of the line and neutral conductors' - this tells you that there are two conductors supplying the load (and it is single phase) and if the the line has a resistance of Rline and the neutral has a resistance of Rneut then Rline + Rneut = 0.45 Ohms.

Why have they mentioned 20 degrees C? This is because the resistance of a conductor depends on it is temperature. For metallic conductors the higher the temperature the higher the resistance and the lower the temperature the lower the resistance. So to tell you something truthful and in fullest detail they give the resistance for the actual temperature of the conductors. Mentioning the temperature with a figure is a way the examiner is testing your understanding of the resistance of conductors and giving a number which you may or may not need to use in the calculation on actual voltage drop. It is the temperature of the conductors while 20Amps is flowing through them both.

Whether the voltage drop takes place equally between line and neutral conductors does not matter; it is the total effect of resistance between the supply and the load which determines the voltage drop.

So, Supply - resistance of line and neutral conductors - load.

Then you are told the current which is drawn by the load which is 20A. This current flows through the load and also through the resistances of the line and neutral conductors. We know that what I have underlined is 0.45 Ohms.

The 'Volt drop' they are asking about is across the resistance of the line and neutral conductors.

So, in your mind , and in a diagram if you drew one - which you should always - you have a resistance of 0.45 Ohm and through it is passing a current of 20 A. How do you calculate the voltage drop across the resistor? Remembering Ohm's Law you know V=I x R.

So Vdrop is 20 x 0.45= 9Volts. Don't forget to put in Volts or V after the number if you wamt full marks.

You do not need to use the temperature in your calculations.

Instead of writing 'Vdrop is 20 x 0.45 = 9Volts' it would be better to write 'Vdrop at 20C is 20 x 0.45 = 9Volts. This is what I thought they meant by 'actual'.

However, the question may be looking for you to consider the Vdrop when the conductors are operating at 70C in which case you would need to correct the resistance figure of 0.45Ohms by multiplying it by 1.2: so 20 x (0.45 x 1.2) = 10.8 Volts at 70C conductor temperature - a worse case Vdrop when the conductor is operating at the maximum design temperature.

One for your lecturer to clarify for you.

PS: In my on-line research for the use of the correction factor '1.2' I turned up this reference: http://electriciancentre.co.uk/how-to/calculate-voltage-drop/