hw17 - Physics 21 Fall, 2004 Solution to HW-17 30 P 26 In...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 21 Fall, 2004 Solution to HW-17 30 P15 Ignoring any mutual inductance, what is the equivalent inductance of two inductors connected (a) in se- ries, (b) in parallel? (a) The same current I flows through two inductors in series, and the voltage across the two inductors is E = E 2 + E 2 = L 1 dI dt + L 2 dI dt = ( L 1 + L 2 ) dI dt , so the inductance is L = L 1 + L 2 . (b) For two inductors in parallel, the current I splits into I 1 and I 2 , and the voltages across the two inductors are equal: E = E 1 = E 2 = L 1 dI 1 dt = L 2 dI 2 dt . We can write E = L dI dt = L µ dI 1 dt + dI 2 dt . Using dI 1 /dt = −E /L 1 and dI 2 /dt = −E /L 2 , we see that E = L ( E /L 1 + E /L 2 ) 1 L = 1 L 1 + 1 L 2 . 30 P20 What is the energy density at the center of a cir- cular loop of wire carrying a 30 A current if the radius of the loop is 28 cm? The magnetic ±eld at the center of a loop of radius R carrying current I is B = µ 0 I/ 2 R , so the energy density is
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/06/2008 for the course PHYS 21 taught by Professor Hickman during the Fall '07 term at Lehigh University .

Ask a homework question - tutors are online