Q120 / 160 · JIPMER 2013 · Physics

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
    $4 R$
  • C
    $\frac{R}{4}$
  • Correct answer
    D
    $\frac{R}{16}$

Solution

$\begin{array}{ll}R=\frac{\rho l}{A} \text { or } R \propto l \\ \therefore & \frac{R_1}{R_2}=\frac{I_1}{I_2}=\frac{L}{L / 4}=4 \\ \text { or } & R_2=\frac{R}{4}\left(\because R_1=R\right)\end{array}$ In parallel combination of such four resistance $\frac{1}{R^{\prime}}=\frac{1}{R_2}+\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_2}$ or $\quad \frac{1}{R^{\prime}}=\frac{1}{R / 4}+\frac{1}{R / 4}+\frac{1}{R / 4}+\frac{1}{R / 4}$ or $\quad \frac{1}{R^{\prime}}=\frac{4}{R}+\frac{4}{R}+\frac{4}{R}+\frac{4}{R}$ or $\quad \frac{1}{R^{\prime}}=\frac{16}{R}$ or $\quad R^{\prime}=R / 16$

Hence, the correct answer is option: 'D'.

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