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Electromagnetic Induction — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsElectromagnetic Induction5 Marks
Question
An LR circuit having a time constant of 50ms is connected with an ideal battery of emf $\in.$ Find the time elapsed before:
  1. The current reaches half its maximum value.
  2. The power dissipated in heat reaches half its maximum value.
  3. The magnetic field energy stored in the circuit reaches half its maximum value.
✓

Answer

$\tau=\frac{\text{L}}{\text{R}}=50 \ \text{ms}=0.05$
  1. $ \ \frac{\text{i}_0}{2}=\text{i}_0\Big(1-\text{e}^{\frac{\text{-t}}{0.06}}\Big)$
$\Rightarrow\frac{1}{2}=1-\text{e}^{\frac{\text{-t}}{0.05}}=\text{e}^{\frac{\text{-t}}{0.05}}=\frac{1}{2}$

$\Rightarrow\text{ln }\text{e}^{\frac{\text{-t}}{0.05}}=\text{ln}^{\frac{1}{2}}$

$\Rightarrow\text{t}=0.05\times0.693 =0.3465 = 34.6 \text{ms}=35 \text{ms}.$
  1. $\text{P}=\text{i}^2\text{R}=\frac{\text{E}^2}{\text{R}}\Big(1-\text{E}{\frac{\text{-t.R}}{\text{L}}}\Big)^2$
Maximum power $=\frac{\text{E}^2}{\text{R}}$

So, $ \frac{\text{E}^2}{2\text{R}}=\frac{\text{E}^2}{\text{R}}\Big(1-\text{e}\frac{\text{-t.R}}{\text{L}}\Big)^2$

$\Rightarrow 1-\text{e}\frac{\text{-tR}}{\text{L}}=\frac{1}{\sqrt2}=0.707$

$\Rightarrow\text{e}\frac{\text{-tR}}{\text{L}}=0.293$

$\Rightarrow\frac{\text{tR}}{\text{L}}=-\text{In} \ 0.239=1.2275$

$\Rightarrow \text{t}=50 \times1.2275 \ \text{ms}=61.2 \text{ms}.$

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