An electric cable has just one wire of radius 9mm,its resistance is 5 ohm. The single wire of the cable is replaced by 6 different well insulated wires each of radius 3 mm. the to tal resistance is 7.5,, but there is a nother option of 270 ohm?
The Q is 7.5 ohm is the ans when the resistances are taken in parallel, but they are not mentioned anywhere??
And the ans is 270 when in series?
Why is the resistance taken in parallel?

It sounds like the analysis is the following:

Given that your 9 mm diamater wire has resitance of 5 ohms, and using an assumption that the resistance of a wire is inversely related to its cross-section, it would seem that a wire made of the same material and 3mm in diamater would have 9 times the resistance (since 3 mm squared is one ninth of 9 mm squared). So one wire of 3 mm diameter would be exprected to have 5*9 = 45 ohms resistance. With me so far? Ok, now if you take a bundle of 6 of these 45-ohm resistances and wire them in parallel then the total resistance of the bundle can be found from:

1/R(total) = 1/R1 + 1/R2 + 1/R3 +... + 1/R6

Solve for R(total) and you'll find it equals 45/6 = 7.5 ohm.

On the other hand if you were to connect all 6 wires in series (end-to-end) you would end up with a long string whose resistance is 6 times 45 ohms, or 270 ohms.

All you really need to remember is that when you wire resistances together in series you add them together, and when you wire them in parallel you use the equation that I used above.

I'll add something. There is the radius and diameter trick usedi n the problem. One references radius and the other diameter.

Wires in parallel act like resistors in parallel. Adding wires in parallel decreases the resistance and it's always lower than the lowest R. When you have two R's of the same vaue in parallel, the value is half. 8 || 8 = 4. An 8 || 3 will be less than 3. Using 1/Rt = 1/R1+1/R2+... +1/Rn

I'll add something. There is the radius and diameter trick usedi n the problem. One references radius and the other diameter.


My bad - I should have said "radius" instead of "diameter." At least I was consistent - so the answer still stands.

No ,I got the answer for both cases, anyway thanks for the answers!!
But this was a multiple choice q and the ans was 7.5, so my Q is how can we just make an assumption that the wires were connected in parallel and not in series! i.e. why were they connected in parallel? Any specific reason , which I can keep in mind for similar Q's
Thanks

Tanuta - I suggest you post the question as it was posed to you - that would help us understand what your confusion is over.

Well the Qu solved was correct , but why was it parallel?
The ans book says it was a parallel connection not series?

I mean the 6 resistances were connected in parallel? Right? Why

When you "replace" a wire, you generally don't make it 5 times longer.

Say you replaced a wire to your stove. You replaced it because it could no longer carry the required current, so you put in a thicker wire in the same path.

If you took out a wall, you may have to make it longer, but not 5x longer.

i mean the 6 resistances were connected in parallel? right? why
Putting 6 wires that are each 45 ohm in parallel is the only way to get the total resistance through the 6 wires down to 7.5 ohm.

Thank you for the answers .they were very helpful indeed!
And I thnk I got the ans.

In Electrical use, you can't parallel Smaller than 1/0, unless it is less than 50 volts.
Like mentioned above, Resistors in parallel formula is The reciprocal of the sum of the reciprocals, it is the same formula for Capacitors in SERIES. May be unnecessary info?

Sorry statmando but could u explain again I did not understand?

1/Rt=1/R1+1/R2+... 1/Rn (resistors in parallel)

1/Ct=1/C1+1/C2.. 1/Cn )capacitors in series)

1/0 is a way of writing 0 AWG
3/0 is a way of writing 000 AWG

By the NEC (National Electric Code: NFPA 70) in the US paralleling cables is not permitted as he states.

Your problem is theoretical. It's also metric.