Ohms Law Quiz

[ban-all-sheds]

OOI, how hard was that supposed to be?

 

[GaryMo]

Different resistances wouldn't make it any harder, it would just create messy numbers.

 

[GaryMo]

OOI, how hard was that supposed to be?

I would like to guess that the majority of electricians would fail that test (not including the ones on here though )

 

[ericmark]

OK so try this one.

Untitled

• ericmark
• 22 Feb 2009
• 1

I did not want any mistakes with answer so copied out of John Bird Electrical Circuit Theory and Technology no cheating answer already in my pictures on this site in case you want to peep.

 

[ban-all-sheds]

I'd hate to think that anybody charging for his services would know so little about his chosen field of expertise that he got it wrong...

 

[ban-all-sheds]

A diagram a tad on the small side for a laptop screen at 1440x900 resolution....

 

[GaryMo]

I'd hate to think that anybody charging for his services would know so little about his chosen field of expertise that he got it wrong...

It's exactly that way though an installation electrician would rarely adds the sum of series and parallel resistors and voltage drop across them since their previous college days.

 

[Spark123]

The one above is slightly more complex as in the current does not flow in the direction which you expect and will go the opposite way through the right hand side cell.
Unless that is if we are measuring voltage using a high impedance meter each one will measure approximately the same as the supply voltage local to it.

 

[ban-all-sheds]

Here's a practical one - I had to work this out once (well - an exact equivalent problem - I may have not remembered the cable lengths correctly).

The use of s/w packages to solve this is forbidden.

There are no tricks - I mean the lamp at the far end of the cable, it's copper cable, normal temperature, the transformer puts out 12V and is adequately rated for 300W etc etc.

 

[Spark123]

Not being bothered to work out the new currents from each lamp owing to their reduced voltages, the VD at the end lamp will be 9.93v rounded hence the voltage is 2.07v.

 

[ricicle]

I got 9.525V at the end lamp (with some rounding up of decimals)

 

[Spark123]

Is that voltage or volt drop? Did you use 18mV/A/m like I did?

 

[ricicle]

Is that voltage or volt drop? Did you use 18mV/A/m like I did?

9.525V across the end lamp. 2.475V drop @ 18mV/A/M

 

[Spark123]

I think I noticed where I went wrong
Now I got a volt drop of 7.35 volts which gives a voltage of 4.65v across end lamp

 

[ericmark]

Total resistance = 0.578989945 = 20.72574851A.
Volt drop first leg = 1.11919042V
second 0.318303533
third 0.25228787
fourth 0.187849007
Fifth 0.124584199
Sixth 0.062486156
leaving 9.935298815 volts at the last lamp.
I can see the problem as the last lamp has less volts so will not draw the same current as first.
Showing all working
2.88 R1
0.018
2.898
1.462485981 R1+R2
0.98794202 R1+2+3
0.753603844 R1+2+3+4
0.615307567 R1+2+3+4+5
0.524989945 R1+2+3+4+5+6
0.578989945 Total
20.72574851 Amps
1.11919042 Volt drop lamp 1
17.68352962 Amps
0.318303533 Volt drop lamp 2
14.01599279 Amps
0.25228787 Volt drop lamp 3
10.43605593 Amps
0.187849007 Volt drop lamp 4
6.921344409 Amps
0.124584199 Volt drop lamp 5
3.471453115 Amps
0.062486156 Volt drop lamp 6
9.935298815 At last lamp
So now lets see what BAS measured it at?