Question
An LR circuit has L = 1.0H and $\text{R}=20\Omega.$ It is connected across an emf of 2.0V at t = 0. Find $\frac{\text{di}}{\text{dt}}$ at:
- t = 100ms
- t = 200ms
- t = 1.0s
✓
Answer
$\text{L}=1.0 \text{H}, \ \text{R}=20 \Omega, \ \text{emf}=2.0\text{V }$$\tau=\frac{\text{L}}{\text{R}}=\frac{1}{20}=0.05 $
$\text{i}_0=\frac{\text{e}}{\text{R}}=\frac{2}{20}=0.1\text{A} $
$\text{i}=\text{i}_0(1-\text{e}^{-\text{t}})=\text{i}_0-\text{i}_0\text{e}^{-\text{t}} $
$\Rightarrow\frac{\text{di}}{\text{dt}}=\frac{\text{di}_0}{\text{dt}}\Big(\text{i}_0\times-\frac{1}{\tau}\times\text{e}^\frac{-\text{t}}{\tau}\Big)=\frac{\text{i}_0}{\tau}\text{e}^\frac{-\text{e}}{\tau }$
So,
$\text{i}_0=\frac{\text{e}}{\text{R}}=\frac{2}{20}=0.1\text{A} $
$\text{i}=\text{i}_0(1-\text{e}^{-\text{t}})=\text{i}_0-\text{i}_0\text{e}^{-\text{t}} $
$\Rightarrow\frac{\text{di}}{\text{dt}}=\frac{\text{di}_0}{\text{dt}}\Big(\text{i}_0\times-\frac{1}{\tau}\times\text{e}^\frac{-\text{t}}{\tau}\Big)=\frac{\text{i}_0}{\tau}\text{e}^\frac{-\text{e}}{\tau }$
So,
- $\text{t}=100\text{ms}\Rightarrow\frac{\text{di}}{\text{dt}}=\frac{0.1}{0.05}\times\text{e}^\frac{-0.1}{0.0.05}=0.27 \text{A}$
- $\text{t}=200\text{ms}\Rightarrow\frac{\text{di}}{\text{dt}}=\frac{0.1}{0.05}\times\text{e}^\frac{-0.2}{0.05}=0.0366 \text{A }$
- $\text{t}=1\text{s}\Rightarrow\frac{\text{di}}{\text{dt}}=\frac{0.1}{0.05}\times\text{e}^\frac{-1}{0.05}=4\times10^{-9}\text{A }$
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