Q133 / 180 · AIIMS 2018 (26 May) · Physics

If current in diode is five times that in $R_1$. Breadkdown voltage of diode is 6 volt. Find the value of $R$. Figure

  • A
    $2000 \Omega$
  • Correct answer
    B
    $2000 / 3 \Omega$
  • C
    $1000 \Omega$
  • D
    $1000 / 3 \Omega$

Solution

Current in diode is 5 times that in $R_1$. Current through $R_1$ is $I_1=\frac{V}{R_1}=\frac{6}{1 \times 10^3} \mathrm{~A}=6 \mathrm{~mA}$ So total current drawn from battery $=6 \mathrm{~mA}+30 \mathrm{~mA}=36 \mathrm{~mA}$ Potential difference across $R=24$ volt So $V=I R ; \quad 24=36 \times 10^{-3} R ; \quad R=2000 / 3 \Omega$

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

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