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mikedunne101

I have some sample questions im revising for my 2330 unit 1 exam. I was hoping somebody wouldnt mind explaining how to do them correctly!

20. 04 02 3
The current in a coil changes from 5A to 2A in 0.05seconds and induces a voltage of 30v in the coil the inductance of the coil is:

22. 04 04 5
A coil has an inductance of 0.1 H and is connected across a 50Hz a.c. supply, the value of inductive reactance of the coil will be:

18. The resistance of a 230v 3 kW immersion heater element will be

19. The force on a conductor 750mm long carrying a current of 6A placed at right angles in a field of 1.2 Tesla will be

21. A capacitor of 100µF is connected in parallel with a 20Ω pure resistor. The group is connected to a 200v 50Hz supply. Calculate the power dissipated in the combination.

sorry i know there is quite a few! any help would be appreciated, cheers.
 
Hi are you doing the first year or second?? sounds like you revising for the unit 3 exam not unit 1 :)


22.A coil has an inductance of 0.1 H and is connected across a 50Hz a.c. supply, the value of inductive reactance of the coil will be:

2 x 3.141 x 5o x 0.1 = 31.41 ohms

so it 2 x π x frequency x inductance
i think thats right anyway :)
 
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18. The resistance of a 230v 3 kW immersion heater element will be

Power = volts squared devided by resistance

P=VxV
......R

3000=230X230
..............R
therefore
. 1 .= 1 R 1..
3000..230X230

therefore
R= 230 X230 =17.63ohms
........3000
 
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formula is e=-L times change in current divided by change in time. Need to transpose. .-05x30 divided by 2-5= .5H. hope you understand that.
 
Hi,go to the nz electrical workers registration board webb site,you will find all the questions and model answers for the electricalns theory papers,the regs paper & inspectors exam also available,not really relevant to uk, but has a few calculation examples as well.

good luck!
 
Hi are you doing the first year or second?? sounds like you revising for the unit 3 exam not unit 1 :)


22.A coil has an inductance of 0.1 H and is connected across a 50Hz a.c. supply, the value of inductive reactance of the coil will be:

2 x 3.141 x 5o x 0.1 = 31.41 ohms

so it 2 x π x frequency x inductance
i think thats right anyway :)



That’s Correct

XL = Inductive Reatance

XL = ωL = 2∏fL

21. A capacitor of 100µF is connected in parallel with a 20Ω pure resistor. The group is connected to a 200v 50Hz supply. Calculate the power dissipated in the combination


Xc = 1/ 2∏FC

= 1/2∏ x50x 100x10-6

= 31.831


Z = √ (R2 +Xc2) (this is R squared + Xc squared)

= 37.59 Ω


I = V /Z

= 200/ 37.59

= 5.32 A


Cos Ø = R/Z

= 20/37.59

= 0.53

PF = 57.99◦


P = V I Cos Ø

= 200 X5.32 X 0.53

= 563.92 W



20. 04 02 3
The current in a coil changes from 5A to 2A in 0.05seconds and induces a voltage of 30v in the coil the inductance of the coil is:

V = L x (di/dt)

30 = L x (3/0.05)

30 = L x 60

L = 30/60

= 0.5 H

= 500mH



Hope this helps been a while so double check please :)
 
Last edited by a moderator:

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