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JIPMER 2013
Exam
JIPMER
Year
2013
Subject
Physics
Chapter
Current Electricity

JIPMER 2013 · Physics · Current Electricity

A uniform wire of resistance R and length L is cut into four equal...

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

Options

  1. ARR
  2. B4R4 R
  3. CR4\frac{R}{4}
  4. DR16\frac{R}{16}

Correct answer

D. R16\frac{R}{16}

Step-by-step solution

R=ρlA or R∝l∴R1R2=I1I2=LL/4=4 or R2=R4(∵R1=R)\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 1R′=1R2+1R1+1R2+1R2\frac{1}{R^{\prime}}=\frac{1}{R_2}+\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_2} or 1R′=1R/4+1R/4+1R/4+1R/4\quad \frac{1}{R^{\prime}}=\frac{1}{R / 4}+\frac{1}{R / 4}+\frac{1}{R / 4}+\frac{1}{R / 4} or 1R′=4R+4R+4R+4R\quad \frac{1}{R^{\prime}}=\frac{4}{R}+\frac{4}{R}+\frac{4}{R}+\frac{4}{R} or 1R′=16R\quad \frac{1}{R^{\prime}}=\frac{16}{R} or $\quad R^{\prime}=R / 16
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