ElectroMagnetic InductionMedium
Question
An inductor of inductance 10 mH having resistance of $100\Omega$ is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is 2 mA and 4 mA is $\_\_\_\_$ .
Options
A.$4/3$
B.$3/4$
C.$5/3$
D.$3/5$
Solution
Sol. [Key Concept: During growth the inductor voltage is $V_{L} = E - iR$; evaluate it at $i = 2$ mA and $i = 4$ mA and take the ratio.]
Apply KVL
$${E - iR - V_{L} = 0 }{V_{L} = E - iR = 1 - 100i }$$at $i_{1} = 2 \times 10^{- 3}\text{ }A$
$$V_{L_{1}} = 0.8\text{ }V $$at $i_{2} = 4 \times 10^{- 3}\text{ }A$
$${V_{L_{2}} = 0.6\text{ }V }{\therefore\frac{V_{L_{1}}}{{\text{ }V}_{L_{2}}} = \frac{4}{3} }$$
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