B) Vindya connects a conductor of length L and thickness R with a source of V volts and she notices the current in the conductor is I amperes. Vasini connects a I conductor of same material of length L1 and radius R1 with the same source and notices the current in the conductor has not changed. If the length of Vasini's conductor is half of Vindya's conductor, What is the radius of Vasini's conductor in terms of Vindyas conductor? Explain.
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Sed Life ikr
I hate her too
she banned my i.d for a week
Answers & Comments
Given:
Case - 1
Conductor 1 connected by Vindya
Length = L, Thickness = Radius = R, Potential difference = V, Electric current = I Ampere
Case - 2
Conductor 2 connected by Vasini
Length = L/2, Thickness = Radius = R1, Potential difference = V, Electric current = I Ampere
Solution:
Potential difference = Electric current × Resistance
or, V = R' (in both cases) ------ eq. (i)
To find radius of Vasini's conductor in terms of Vindya's conductor:
In conductor - 1,
Resistivity (ρ₁) = R' ₓ A/L (Area is the thickness of wire i.e R)
or, ρ₁ = VR/L
In conductor - 2,
Resistivity (ρ₂) = R' ₓ A1/L/2
or, ρ₂ = R' ₓ 2/A1 ₓ L
or, ρ₂ = 2V/R1 L ⇒From eq. (i) (A1 = R1)
Since resistivity is defined in terms of resistance (R')
ρ₁ = ρ₂
VR/L = 2V / R1 ₓ L
or, R/R1 = 1
Thus, the radius of Vasini's conductor in terms of Vindya's conductor is
R/R1 = 1
Step 1: Given data
Conductor 1 is connected by Vindya having,
Length[tex]=L[/tex],
Thickness [tex]=[/tex] Radius [tex]=R[/tex] ,
Potential difference [tex]=V[/tex],
Electric current [tex]=I[/tex]
Conductor 2 is connected by Vasini having,
Length[tex]=\frac{L}{2}[/tex],
Thickness [tex]=[/tex] Radius[tex]=R_{1}[/tex],
Potential difference [tex]=V[/tex],
Electric current [tex]=I[/tex]
Step 2: To find the radius of Vasini's conductor in terms of Vindya's conductor
We know that,
Potential difference [tex]=[/tex] Electric current [tex]\times[/tex] Resistance
i.e., [tex]V=I\times R[/tex]
[tex]R=\frac{\rho\times L}{A}[/tex]
[tex]\therefore \rho=\frac{R\times A}{L}[/tex]
or [tex]\rho=\frac{V\times A}{I \times L}[/tex]
Since area[tex]=[/tex]thickness[tex]=[/tex]radius,
[tex]\therefore A=R[/tex] and [tex]A1 = R1[/tex]
In conductor - 1,
Resistivity [tex]\rho_{1} = R' A /IL[/tex] (Area is the thickness of wire i.e [tex]R[/tex])
or, [tex]\rho_{1} = VR /IL---(1)[/tex]
In conductor - 2,
Resistivity [tex]\rho_{2} = R' A_{1}/I(L/2)[/tex]
or, [tex]\rho_{2} = 2R' A_{1}/IL[/tex]
or, [tex]\rho_{1} = 2VR_{1} /IL---(2)[/tex]
From (1) and (2),
[tex]VR /IL= 2VR_{1}/IL\\R=R_{1}[/tex]
Hence, the radius of Vasini's conductor in terms of Vindya's conductor is [tex]R_{1}=R[/tex].
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