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Electromagnetic Induction — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsElectromagnetic Induction5 Marks
Question
Two coils A and B have inductances 1.0H and 2.0H respectively. The resistance of each coil is $10\Omega.$ Each coil is connected to an ideal battery of emf $2.0V$ at $t = 0$ Let $i_A$ and $i_B$ be the currents in the two circuit at time t. Find the ratio $\frac{\text{i}_\text{A}}{\text{i}_\text{B}}$
  1. t = 100ms
  2. t = 200ms
  3. t = 1s.
✓

Answer

$\text{L}_\text{a}=1.0\text{ H }; \text{ L}_\text{B}=2.0\text{H}; \ \text{R}=10\omega$
  1. $\text{t}=0.1\text{s}, \ \tau_\text{A}=0.1, \ \tau_\text{B}=\frac{\text{L}}{\text{R}}=0.2$
$\text{i}_{\text{A}}=\text{i}_0\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)$
$=\frac{2}{10}\Big(1-\text{e}^\frac{-0.1\times10}{1}\Big)=0.2\Big(1-\text{e}^{-1}\Big)=0.126424111$
$\text{i}_\text{B}=\text{i}_0\Big(1-\text{e}\frac{-\text{t}}{\tau}\Big)$
$=\frac{2}{10}\Big(1-\text{e}^\frac{-0.1\times10}{2}\Big)=0.2\Big(1-\text{e}^\frac{-1}{2}\Big)=0.078693$
$\frac{\text{i}_\text{A}}{\text{i}_\text{B}}=\frac{0.12642411}{0.78693}=1.6$
  1. $\text{t}=200\text{ms}=0.2\text{s}$
$\text{i}_\text{A}=\text{i}_0\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)$
$=0.2\Big(1-\text{e}^\frac{-0.2\times10}{1}\Big)=0.2\times0.864664716=0.172932943$
$\text{i}_\text{B}=0.2\Big(1-\text{e}^\frac{-0.2\times10}{2}\Big)=0.2\times0.632120=0.126424111$
$\frac{\text{i}_\text{a}}{\text{i}_\text{B}}=\frac{0.172932943}{0.126424111}=1.36=1.4$
  1. $\text{t}=1\text{s}$
$\text{i}_\text{A}=0.2\Big(1-\text{e}^\frac{-1\times10}{1}\Big)=0.2\times0.9999546=0.19999092$
$\text{i}_\text{B}=0.2\Big(1-\text{e}^\frac{-1\times10}{2}\Big)=0.2\times0.99326=0.19865241$
$\frac{\text{i}_\text{A}}{\text{i}_\text{B}}=\frac{0.19999092}{19865241}=1.0$

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