901_PartUniversity Physics Solution

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

This preview shows pages 1–2. Sign up to view the full content.

Inductance 30-25 (b) Using m 1 K χ =+ we can find the inductance for any height 0m 1 d LL D ⎛⎞ ⎜⎟ ⎝⎠ . _______________________________________________________________________________ Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury 4 dD = 0.63024 H 0.63000 H 2 = 0.63048 H 0.62999 H 34 = 0.63072 H 0.62999 H = 0.63096 H 0.62998 H ________________________________________________________________________________ The values 35 m2 m (O ) 1.52 10 and (Hg) 2.9 10 χχ −− = × have been used. EVALUATE: (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. IDENTIFY: 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. SET UP: The equivalent inductance is defined by eq di L dt = E , where E and i are the total emf and current across the combination. EXECUTE: Series: 1 212 1 2 21 12 eq . di di di di di M M L dt dt dt dt dt ++ + But 12 1 2 12 21 and di di di iii M M M dt dt dt =+⇒ = + = , so e q (2 ) di di LL M L dt dt = and eq 1 2 2 L =++ . Parallel: We have 11 2e q di di di LM L dt dt dt += and 21 22 1 e q di di di L dt dt dt , with di di di dt dt dt + = and 12 21 M MM =≡ . To simplify the algebra let , , and . di di di AB C dt dt dt == = So 1e q q , , . L AM B LCLBM ALCABC + = + = Now solve for and in terms of . C () 0 u s i n g L MA M LB A C B +− = = . ( ) 0 LMCB MLB −+ − = . 2 0 LMC LMBMLB −+ −= . 1 ) ( ) M LLB MLC and 1 . ) ML B C MLL = But 2 1 ()( 2 ) , (2 ) ) M LC M L L M L ACBC C + =−=− = or 2 2 A C M = . Substitute A in B back into original equation: 1 eq 2( 2 ) LM LC MM L CLC and 2 eq . 2 L = Finally, 2 eq 2 M L L = . EVALUATE: If the flux of one coil doesn’t pass through the other coil, so 0 M = , then the results reduce to those of problem 30.47. 30.79. IDENTIFY: Apply Kirchhoff&s loop rule to the top and bottom branches of the circuit. SET UP: Just after the switch is closed the current through the inductor is zero and the charge on the capacitor is zero.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

### Page1 / 5

901_PartUniversity Physics Solution - Inductance 30-25 d...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online