# Electricity question

Watch
Announcements
#1
Hi, Can anyone help me with this one?

A 110v DC generator supplies a 1000W floodlight through 50 metres of two core cable. If voltage at lamp is to be no less than 105v-
A. What is current flowing?
B. What is total power loss in cable?
C. What is maximum resistance per metre of each core of cable?

So far I have for Part A-

Power = V x I
1000 = 110 x I
I = 1000 / 110
So Current = 9.09A

Power = V2/R
R =V2/ Power
= 110x110/1000
= 12.1 Ohms

Part B

Power drop = W=I2/R
=9.09x9.09x12.1
=999.8 Watts per cable
= 1999.6 Watts in total

And for Part C

12.1 Ohms divided by 100 metres of cable = 0.121 Ohms per metre

I'm unsure if I have gone about this the right way, and also where the voltage difference (i.e. 110 v supply, but 105 v at lamp) will come into the answer.

Any help would be very gratefully received!!

Thanks
Murph 0
6 years ago
#2
(Original post by murphymurphy)
A 110v DC generator supplies a 1000W floodlight through 50 metres of two core cable. If voltage at lamp is to be no less than 105v-
(Original post by murphymurphy)
A. What is the current flowing?

So far I have for Part A-

Power = V x I
1000 = 110 x I
I = 1000 / 110
So Current = 9.09A
Hello and welcome to TSR.

Looking at your attempted answers, there is one tip that will help enormously: Always draw a circuit. It will help clarify how to approach the question.

You have the correct formula but your understanding of the question needs a little help.

The circuit comprises a 110V dc source, connected via a (twin core 2x50m length conductor) cable of unknown resistance (Rcableohms), to the lamp load which can be thought of as another resistance (RLamp ohms).

i.e. a potential divider with the cable represented by one resistance and the lamp load as the other.

The question also states that the pd across the lamp must be no less than 105V dc.

This is saying that a pd of 5 Volts is dropped across the resistance of the cable. i.e. the total current through the cable together with the cables resistance will produce a power loss.

If the lamp is to provide full power output of 1000W, then the cable power losses must be in addition to this.

So the power equation P= VI must use VLamp and PLamp in order to find the correct current flowing through the lamp to meet this specification.

i.e. Plamp = 1000W and Vlamp = 105V to find Ilamp

Spoiler:
Show
PLamp = Vlamp x Ilamp

Ilamp = Plamp / Vlamp

Ilamp = 1000 / 105

Ilamp = 9.524 Amperes (3 dp)

(Original post by murphymurphy)
B. What is the total power loss in the cable?

Power = V2/R
R =V2/ Power
= 110x110/1000
= 12.1 Ohms

Part B

Power drop = W=I2/R
=9.09x9.09x12.1
=999.8 Watts per cable
= 1999.6 Watts in total
The cable has a finite unknown resistance. But we know that 1000W is dissipated in the lamp and some power must also be dissipated in that cable resistance.

Notice also that the cable is in series with the lamp so the lamp current must be the same as the current flowing through the cable resistance.

i.e. Icable = Ilamp

You can use either P = I2R or P = V2/R

Where:

V is the pd across the cable i.e. 110V - 105V = 5V

I is the current calculated in part (a) = 9.524 A

And R is the cable resistance which can be found from R = V/I

Spoiler:
Show
Rcable = Vcable / Icable

Rcable = (110 - 105) / 9.524

Rcable = 5V / 9.524A

Rcable = 0.525 ohms (3 dp)

Then using:

Ploss = V2/R where V = pd across the cable and R is the cable resistance

Ploss = Vcable / Rcable

Ploss = 52 / 0.525

Ploss = 25 / 0.525

Ploss = 47.62 Watts (2 dp)

(Original post by murphymurphy)
C. What is maximum resistance per metre of each core of cable?

And for Part C

12.1 Ohms divided by 100 metres of cable = 0.121 Ohms per metre

I'm unsure if I have gone about this the right way, and also where the voltage difference (i.e. 110 v supply, but 105 v at lamp) will come into the answer.
Use the cable resistance calculated in part (b) above.

Spoiler:
Show
Rcable / length = 0.525ohms / (2 x 50) metres

0.525 / 100 = 0.00525 ohms/m

That is, the maximum conductor core resistance per metre allowed to achieve the 1000W power delivered to the load if the pd across the load is to be not less than 105V is:

5.25x10-3 ohms/m
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (226)
61.08%
No (77)
20.81%
Not sure (67)
18.11%