L=101 1r2(P4() dt q 2 L=10l Q1n4() dt ec 2v1                =  dt p t p t t A 2 1 2 0 2 2 1 eq ( ) t -t 1 L 10lg       =  dt t t L t p A 2 1 0.1 ( ) 2 1 eq 10 t -t 1 L 10lg