Two different coils have self-inductances L₁ = 8 mH and L₂ = 2 mH. At a certain instant, the current in the two coils is increasing at the same constant rate and the power supplied to the two coil is the same. Find the ratio of (a) induced voltage (b) current and (c) energy stored in the two coils at that instant?

Solution

Given, two different coils.
L₁ = 8 mH; L₂ = 2 mH

a) Now, using the formula,
$e = L \frac{dI}{dt} \Rightarrow \frac{e_{1}}{e_{2}} = \frac{L_{1}}{L_{2}} = \frac{8}{2} = 4$

Hence, ratio of induced volatge is 4:1 .

b) Since, $P = eI = constant$

Therefore, $\frac{{dI}_{1}}{dt} = \frac{{dI}_{2}}{dt}$

$P_{1} = P_{2} = P$

$e_{1} I_{1} = e_{2} I_{2}$

$∴$ $\frac{I_{1}}{I_{2}} = \frac{e_{2}}{e_{1}} = \frac{1}{4}$
Ratio of current is 1:4 .

c) Energy is given by, $U = \frac{1}{2} {LI}^{2}$

$∴$ $\frac{U_{1}}{U_{2}} = \frac{\frac{1}{2} L_{1}}{\frac{1}{2} L_{2}} {(\frac{I_{1}}{I_{2}})}^{2} = \frac{8}{2} {(\frac{1}{4})}^{2} = \frac{1}{4} .$
Thus ratio of energy is 1 :4 .

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