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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 discharging R-C circuit is given by $\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$ where $i_0$, R and C are constant parameters and t is time. Let $\text{i}_0=2.00\text{A},\text{R}=600\times10^5\Omega$ and $\text{C}=0.500\mu\text{F}.$
  1. Find the current at t = 0.3s.
  2. Find the rate of change of current at t = 0.3s.
  3. Find approximately the current at t = 0.31s.
✓

Answer

Equation $\text{i = i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$$\text{i}_0=2\text{A, R}=6\times10^{-5}\Omega,\text{C}=0.0500\times10^{-6}\text{F}=5\times10^{-7}\text{F}$
  1. $\text{i}=2\times\text{e}^{\Big(\frac{-0.3}{6\times0^3\times5\times10^{-7}}\Big)}=2\times\text{e}^{\big(\frac{-0.3}{0.3}\big)}=\frac{2}{\text{e}}\text{amp}$
  2. $\frac{\text{di}}{\text{dt}}=\frac{-\text{i}_0}{\text{RC}}\text{e}^{-\frac{\text{t}}{\text{RC}}}$ when t = 0.3
$\sec\Rightarrow\frac{\text{di}}{\text{dt}}=-\frac{2}{0.30}\text{e}^{\big(\frac{-03}{0.3}\big)}=\frac{-20}{3\text{e}}\text{amp/sec}$
  1. At $\text{t}=0.31\text{ sec},\text{i}=2\text{e}^{\big(-\frac{-0.3}{0.3}\big)}=\frac{5.8}{3\text{e}}\text{amp}.$

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