The ratio of time constant during current growth and current decay of the circuit shown in figure is

1 : 1

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3 : 2

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2 : 3

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1 : 3

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Solution

The correct option is B $3 : 2$
during charging $L_{1} = \frac{L}{2 R}$ .....(i)

during dischanrging $L_{2} = \frac{L}{3 R}$ .....(ii)

from (i) and (ii)

$\frac{L_{1}}{L_{2}} = \frac{\frac{L}{2 R}}{\frac{L}{3 R}} = \frac{3}{2}$
$\to 3 : 2$

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