In the circuit, a battery of emf E, a resistance R and inductance coil `L_(1) and L_(2)` and switch S are connected as shown. In Fig. Initially the switch is open

Options

A. The time constant of the circuit is `1/R (L_(1)L_(2))/(L_(1)+L_(2))`
B. Steady state current in the inductor `L_(1) is `(EL_(2))/(R(L_(1)+L_(2)))`
C. Steady state current in the inductor `L_(2)` is `(EL_(1))/(R(L_(1)+L_(2)))`
D. In steady the total energy stored in the inductor coils is `1/2 (L_(1)L_(2))/(L_(1)+L_(2)) (E^2)/(R^2)`

Answer

A, B, C, D

Explanation

At any time: `L_(1)(di_(1))/(dt) =L_(2)(di_(2))/(dt)`
`implies int_(0)^(i_(1)) L_(1)di_(1)=int_(0)^(i_2) L_(2)di_(2) implies L_(1)i_(1)=L_(2)i_(2) implies (i_1)/(i_2)=(L_2)/(L_1)` .. (i)
Total steady state current
`i=E//R=i_(1)+i_(2)` ..(ii)
From equation (i) and (ii)
`i_(1)=(EL_(2))/(R(L_(1)+L_(2))), i_(2)=(EL_(1))/(R(L_(1)+L_(2)))`
`U=1/2 L_(eq) i^(2)=1/2 ((L_(1)L_(2))/(L_(1)+L_(2))) (E^2)/(R^2)`.