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Physics and Mathematics — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsPhysics and Mathematics2 Marks
Question
The electric current in a charging R-C circuit is given by $\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$ where $\text{i}_0,$ R and C are constant parameters of the circuit and t is time. Find the rate of change of current at
  1. t = 0.
  2. t = RC.
  3. t = 10RC.
✓

Answer

Given that, $\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$$\therefore$ Rate of change of current $=\frac{\text{di}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\text{i}_0\text{e}^{-\frac{\text{i}}{\text{RC}}}=\text{i}_0\frac{\text{d}}{\text{dt}}\text{e}^{-\frac{\text{t}}{\text{RC}}}=\frac{-\text{i}_0}{\text{RC}}\times\text{e}^{-\frac{\text{t}}{\text{RC}}}$
  1. When $\text{t}=0,\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RC}}$
  2. when $\text{t}=\text{RC},\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RCe}}$
  3. when $\text{t}=10\text{RC},\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RCe}^{10}}$

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