92 Dr. B. van der Pol : Value of Conductivity of Sea- Water 

 have to satisfy the equations 



T dii ^ . 1 «,. . 



L<i -j- +Jct 1 i 1 + p \(ii + i a )dt= — -E cos&)£, 



2 dt + 2 * 2 + C K*i + *a)^ = 0, 



(1) 



where B cos cot is the impressed E.M.F. For simplification 

 we take two special cases ; when 



A. Li = L 2 and under resonance condition 



(LA=L 2 C 2 =«- 2 ) 



the solutions are, leaving out the transients, 



CR 



T-T- • ±Li COS &)£, 



H —n 1 R 2 G+L 1 - :Ecoi(0t > 



1 



(O 



y • ■ ■ (2) 



R 1 R 2 ^ + L 1 J 



B. I£, on the other hand, when the shunt has no appreciable 

 inductance, L 2 in (1) is assumed to be zero, we get the 

 following expression for the currents under resonance 

 condition : 



[(Lx + CRxiy + L 1 rr- pj 2 -i 



R 2 C *• cos cot + ™p- sin cot I 

 A A J 



R x RxRsC + Lx 



t 3 = E ^_coso)i + 



L- A 



}■ (3) 



A A J 



Wh6re A=tl4+0W+^ 



ft) -»• 



When now the resistance Rx in the main branch of the 

 oscillatory circuit is decreased till it finally reaches the ideal 

 value R x = 0, w e see that in both cases A : L 2 = L x and B : L 2 = 

 the current t 2 in the shunt circuit reaches the asymptotic 

 value 



• E • , 

 t 2 = — j — sin cot, 



&) 1^1 



being independent of the resistance R 2 of the shunt across 

 the eondenser. 



