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901_PartUniversity Physics Solution

# 901_PartUniversity Physics Solution - Inductance 30-25 d(b...

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Inductance 30-25 (b) Using m 1 K χ = + we can find the inductance for any height 0 m 1 d L L D χ = + . _______________________________________________________________________________ Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury 4 d D = 0.63024 H 0.63000 H 2 d D = 0.63048 H 0.62999 H 3 4 d D = 0.63072 H 0.62999 H d D = 0.63096 H 0.62998 H ________________________________________________________________________________ The values 3 5 m 2 m (O ) 1.52 10 and (Hg) 2.9 10 χ χ = × = − × have been used. E VALUATE : (d) The volume gauge is much better for the liquid oxygen than the mercury because there is an easily detectable spread of values for the liquid oxygen, but not for the mercury. 30.78. I DENTIFY : The induced emf across the two coils is due to both the self-inductance of each and the mutual inductance of the pair of coils. S ET U P : The equivalent inductance is defined by eq di L dt = E , where E and i are the total emf and current across the combination. E XECUTE : Series: 1 2 1 2 1 2 21 12 eq . di di di di di L L M M L dt dt dt dt dt + + + But 1 2 1 2 12 21 and di di di i i i M M M dt dt dt = + = + = , so 1 2 eq ( 2 ) di di L L M L dt dt + + = and eq 1 2 2 L L L M = + + . Parallel: We have 1 2 1 12 eq di di di L M L dt dt dt + = and 2 1 2 21 eq di di di L M L dt dt dt + = , with 1 2 di di di dt dt dt + = and 12 21 M M M = . To simplify the algebra let 1 2 , , and . di di di A B C dt dt dt = = = So 1 eq 2 eq , , . L A MB L C L B MA L C A B C + = + = + = Now solve for and in terms of . A B C 1 2 ( ) ( ) 0 using L M A M L B A C B + = = . 1 2 ( )( ) ( ) 0 L M C B M L B + = . 1 1 2 ( ) ( ) ( ) 0 L M C L M B M L B + = . 1 2 1 (2 ) ( ) M L L B M L C = and 1 1 2 ( ) . (2 ) M L B C M L L = But 1 1 2 1 1 2 1 2 ( ) (2 ) , (2 ) (2 ) M L C M L L M L A C B C C M L L M L L + = = = or 2 1 2 2 M L A C M L L = . Substitute A in B back into original equation: 1 2 1 eq 1 2 1 2 ( ) ( ) 2 (2 ) L M L C M M L C L C M L L M L L + = and 2 1 2 eq 1 2 . 2 M L L C L C M L L = Finally, 2 1 2 eq 1 2 2 L L M L L L M = + .

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