Question

You are given a wire 1 m long. In which one of the following cases , the energy drawn from the battery will be the largest. Assume that the internal resistance of the battery is zero.

a) Full length of the wire connected across the battery.

b) Half length of the wire connected across the battery.

c) Wire cut into four equal parts and they are connected in parallel to each other across the battery.

d) Wire is cut into two equal parts and they are connected in parallel to each other across the battery.

Answer 1

i think when full length of wire connected across the battery

Answer 2

no ,the answer is not option A)

Answer 3

@Festinger help me

Answer 4

you know the answer>>>>

Answer 5

i can help, but from previous question(s) I haven't been able to make you understand...
The answer is C.

Answer 6

can you explain me that why option C is right? @Festinger

Answer 7

i can surely understand

Answer 8

@Festinger explain me. please

Answer 9

What is power dissipated equal to?

Answer 10

i^2*r

Answer 11

four currents now we have

Answer 12

\[\huge P=VI=\frac{V^2}{R}=I^2R\]

Answer 13

nice. how did you type out in that huge font? i'll go through the derivation now.

Answer 14

or resistance now is less in parallel

Answer 15

use this. @Festinger `huge VI` in equation editor

Answer 16

@needforspeed I get it. Is C.) correct because more current can be withdrawn from the battery since the circuit is in parallel?

Answer 17

The EMF is the same through out the experiments.

In P=VI, V=EMF is constant. Thus, if I is bigger, P is bigger and if I is smaller, P is smaller.

So now what the question wants is the condition at which the power is the greatest, which means it wants you to find the configuration to give the greatest current.

You may recall that if you connect resistors in parallel, the resultant resistance is ALWAYS lower than the lowest resistor connected in parallel. In the special case where the resistors all have the same resistance, the resistance becomes:
\[\frac{1}{R_total}=\frac{1}{R}+\frac{1}{R}+...\]
and thus
\[R_{total}=\frac{R}{n}\]
where n is the number of branches with the same resistance.
Notice that if I connected 2 10Ω reisistor in parallel the total resistance is 5Ω. And If i connected 4 10Ω resistors in parallel the resistance becomes 2.5Ω.

With this logic you should be able to quickly get the answer as C. But if you prefer the math to do the talking...

Assume that the resistance of the 1m wire is R. You can give R a value like 10Ω if it helps you see it.

So if I connected it in series it is P=VI=I^2 R

If I cut the wire into 2, the resistance is halved (using law of proportions) and power is P=4VI^2 R

What if I cut into 2 equal parts and connect them in parallel? Using the argument i mentioned aboved for same resistance resistor connected in parallel:
\[R_{total}=\frac{\frac{R}{2}}{2}=\frac{R}{4}\]

Putting this back into V=IR and P=I^2 R, P= 16V^2 R

What if I split the wire into 4 parts and connected them in parallel?! Ressitance will be even smaller, and current will be even more! i will expect more heat!

\[R_{total}=\frac{\frac{R}{4}}{4}=\frac{R}{16}\]
And
\[P=\frac{256V^{2}}{R}\]

Which has the highest value of them all.

Answer 18

@genius12 you want to say that in parallel combination , the voltage remains constant but current doesn't remain constant , so current is changing and more current will drawn from the circuit.

Answer 19

am i right? @genius12

Answer 20

nice explanation..... i completely understand. thank you.. @Festinger you are awesome

Answer 21

@genius12 tell me am i right?

Answer 22

@Festinger Is EMF electromotive force or is it electric/magnetic fields?

Answer 23

EMF is electromotive force. I equated V=EMF

Answer 24

Oh ok.