Question
An inductor-coil of inductance 20mH having resistance $10\Omega$ is joined to an ideal battery of emf 5.0V. Find the rate of chenge of the induced emf at:
- t = 0
- t = 10ms
- t = 1.0s.
✓
Answer
$\text{L}=20\text{mH} ; \ \text{e}=5.0\text{V},\text{R}=10 \ \Omega$$\tau=\frac{\text{L}}{\text{R}}=\frac{20\times10^{-3}}{10},\text{i}_0=\frac{5}{10}$
$\text{i}=\text{i}_0(1-\text{e}^{\frac{\text{t}}{\tau}})^2$
$\Rightarrow \ \text{i}=\text{i}_0-\text{i}_0\text{e}^{\frac{\text{-t}}{\tau^2}}$
$\Rightarrow\text{iR}=\text{i}_0\text{R}-\text{i}_0\text{R}\text{e}^{\frac{\text{-t}}{\tau^2}}$
$\frac{\text{dE}}{\text{dt}}=10\times\frac{5}{10}\times\frac{10}{20\times10^{-3}}\times\text{e}^{-0.01\times\frac{2}{10^{-2}}}$
$=16.844=17\text{V}/'$
$\text{i}=\text{i}_0(1-\text{e}^{\frac{\text{t}}{\tau}})^2$
$\Rightarrow \ \text{i}=\text{i}_0-\text{i}_0\text{e}^{\frac{\text{-t}}{\tau^2}}$
$\Rightarrow\text{iR}=\text{i}_0\text{R}-\text{i}_0\text{R}\text{e}^{\frac{\text{-t}}{\tau^2}}$
- $10\times\frac{\text{di}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\text{i}_0\text{R}+10\times\frac{5}{10}\times\frac{10}{20\times10^{-3}}\times\text{e}^{0\times\frac{10}{2\times10^{-2}}}$
- $\frac{\text{Rdi}}{\text{dt}}=\text{R}\times\text{i}^0\times\frac{1}{\tau}\times\text{e}^{\frac{\text{-t}}{\tau}}$
$\frac{\text{dE}}{\text{dt}}=10\times\frac{5}{10}\times\frac{10}{20\times10^{-3}}\times\text{e}^{-0.01\times\frac{2}{10^{-2}}}$
$=16.844=17\text{V}/'$
- $\text{For} \ \text{t} =1\text{s}$
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