Solution:
Solution
In steady state when switch was closed,
$${i}_{0}=E/R=(1/5)A=0.2A$$
After switch is opened, it becomes $$L-C$$ circuit in which peak value current is $$0.2A$$
$$\therefore$$ $$\cfrac { 1 }{ 2 } L{ i }_{ 0 }^{ 2 }\quad =\cfrac { 1 }{ 2 } L{ V }_{ 0 }^{ 2 }\quad $$
or $$L=\cfrac{{ V }_{ 0 }^{ 2 }}{{ i }_{ 0 }^{ 2 }}\cdot C$$
$$=\cfrac { { \left( 150 \right) }^{ 2 } }{ { \left( 0.2 \right) }^{ 2 } } \times 0.5\times { 10 }^{ -6 }= 0.28 H$$
