A potential difference of 2 V across a device drives a current of 40 mA through it. When a potential difference of 4 V is applied in the reverse direction, the current is 0.4 mA. The ratio of the resistance \( (V/I) \) of the device in the reverse direction to that in the forward direction is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 200
The device does not obey Ohm’s law, but its resistance at each point is still \( V/I \). The voltages in the two directions are different here. So each resistance must be found on its own.
Forward direction:
\( \begin{aligned} R_f &= \frac{V_f}{I_f} \\ &= \frac{2}{40\times10^{-3}} \\ &= 50\ \Omega \end{aligned} \)
Reverse direction:
\( \begin{aligned} R_r &= \frac{V_r}{I_r} \\ &= \frac{4}{0.4\times10^{-3}} \\ &= 10^{4}\ \Omega \end{aligned} \)
\( \begin{aligned} \frac{R_r}{R_f} &= \frac{10^{4}}{50} \\ &= 200 \end{aligned} \)
Therefore, the resistance of the device in the reverse direction is 200 times its resistance in the forward direction.
NCERT (2026-27):
Points to Ponder – “… The assertion that V = IR is a statement of Ohm’s law is not true. This equation defines resistance and it may be applied to all conducting devices whether they obey Ohm’s law or not. …”