A uniform wire of resistance R and length L is cut into four equal parts, each of length L/4, which are then connected in parallel combination. The effective resistance of the combination will be

R

4 R

$\frac{R}{4}$

$\frac{R}{16}$

Solution

$\frac{R}{16}$

According to ohm's law

R = $\frac{ρ l}{A}$

ρ is a constant of proportionality and depends on the material of a conductor but not on its dimensions. ρ is called resistivity

R ∝ l

i.e resistance is proportional to length

∴ $\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}}$

$= \frac{L}{\frac{L}{4}}$

= 4

$R_{2} = \frac{R}{4}$ ....( $∵$ R₁ = R )

In parallel combination of such four resistances

$\frac{1}{R'} = \frac{1}{R_{2}} + \frac{1}{R_{2}} + \frac{1}{R_{2}} + \frac{1}{R_{2}}$

⇒ $\frac{1}{R'} = \frac{1}{\frac{R}{4}} + \frac{1}{\frac{R}{4}} + \frac{1}{\frac{R}{4}} + \frac{1}{\frac{R}{4}}$

⇒ $\frac{1}{R'} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R}$

⇒ $\frac{1}{R'} = \frac{16}{R}$

⇒ $R' = \frac{R}{16}$

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