Question

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

• A. $R$
• B. $4R$
• C. $\frac{R}{4}$
• D. $\frac{R}{16}$

Answer 1

The correct option is D $\frac{R}{16}$

Resistance of a wire $=\frac{\rho l}{A}$

where, $\rho$ is the resistivity of material of wire

$l$ is length of wire

$A$ is the cross sectional area of the wire.

$\Rightarrow R=\frac{\rho l}{A}$

$\Rightarrow \frac{\rho \left(L/4\right)}{A}=\frac{R}{4}$

Thus, the four smaller wires have resistance $R/4$ each.

Now, these four resistances are connected in parallel.

Thus, effective resistance $R_{eff}$ is given by

$\Rightarrow \frac{1}{R_{eff}}=\frac{1}{R/4}+\frac{1}{R/4}+\frac{1}{R/4}+\frac{1}{R/4}$

$\Rightarrow R_{off}=\frac{R}{16}$