In the given circuit, find the current through the `5 mH` inductor in steady state.

Options

A. Heat flow through slabs A and E is same
B. Heat flow through slab E is maximum
C. Temperature difference across slab E is smallest
D. Heat flow through C = heat flow through B + heat flow through D

Answer

ACD

Explanation

(A) At steady state, heat flow through A and E are same. So option (A) is correct and option (B) is incorrect.
(C ) `DeltaT=HxxR`, where H is heat current, `R=(L)/(KA)`
'H' is same for A and E but R is smallest for E. So temperature difference across slab E is smallest. So option (C) is correct.
(D) `H_(B)=(DeltaT)/(R_(B)),H_(C)=(DeltaT)/(R_(C))andH_(D)=(DeltaT)/(R_(D))`, From this option if . `H_(C)=H_(B)+H_(D)`
`rArr(1)/(R_(C))=(1)/(R_(B))+(1)/(R_(D))rArr(4K(2Lb))/(4L)=(3K(Lb))/(4L)+(5K(Lb))/(4L)`
which is true. Here b is the width of slab. Hence option (D) is also correct.