Question
Question: In how many parts (equal) does a wire of \(100\Omega \) be cut so that a resistance of \(1\Omega \) ...
In how many parts (equal) does a wire of 100Ω be cut so that a resistance of 1Ω is obtained by connecting them in parallel?
A) 10
B) 5
C) 100
D) 50
Solution
As length of the conductor increases, resistance of the conductor also increases and as area of cross section of the conductor increases then resistance decreases. Resistance is the opposition occurs because atoms and molecules of the substance obstruct the flow of charge carriers.
Complete step by step answer:
We know that the resistance of the conductor is directly proportional to the length of the conductor and inversely proportional to the area of cross section.
Then, R∝Al
After removing proportionality, we get, one constant that is ρ
R=ρAl
Where, ρ is the resistivity of the conductor.
Now here, a resistance of a wire 100Ω is divided into equal parts (half) so that we get total resistance as 1Ω when it is connected in parallel. We need to find the total number of wires connected in parallel.
The resistance of one wire of lengthl, R=100Ω (original condition)
Now if we cut it in a half that is equal parts, then
1R1=21R2
Rearranging the above equation, then
R2=2R1
If we cut it in n- equal parts ten resistance will be,
Rn=nR…………(1)
Because, we shorten the length n-times so it becamen1
We haveRp=1Ω then each equal divided wires are connected in parallel, then we get
Rp1=R11+R21+...........+Rn1
In this problem all the resistance connected in parallel are having the same value of resistance. Then
Rp1=R1+R1+...........+R1
11=R1+R1+...........+R1
1=Rnn
Substitute equation (1) in the above equation, we get
1=nRn
1=Rn2
We haveR=100Ω then
1=100n2
⇒n=10
∴ A resistance of 1ohmis obtained by connecting 10 equal parts of the wire connecting them in parallel.
So, option a is correct
Note: Resistance of a conductor depends on the length, area of cross section, temperature of the conductor.
Resistance is the opposition offered by the substance to the flow of current.