ELECTRIC OSCILLATIONS AND ELECTRIC WAVES. 265 



switched on to the line. What is the current and voltage dis- 

 tribution over the line 0.0002 second after the generator is dis- 

 connected? Ans. 10,000 volts opposite to generator voltage, 

 and 1.835 amperes opposite to current which first starts in the line. 



9. The 1 8.6 mile line in problem 5 has connected across its 

 distant end a non-inductive circuit of which the resistance is 

 1,826 ohms. The 10,000 volt generator is connected for 0.0015 

 second. What is the voltage and current distribution over the 

 line? Ans. 10,000 volts and 1.835 amperes over the half of 

 the line next to the generator; and 5,000 volts and 2.752 amperes 

 over the other half of the line. 



10. The distant end of the 18.6 mile line in problem 5 has 

 connected across it a non-inductive circuit of which the resistance 

 is 16,431 ohms. The 10,000 volt generator is connected to the 

 line for o.oooio second and then disconnected. What is the 

 voltage and current distribution over the line 0.00005 second 

 after the generator is disconnected? Ans. Zero voltage and 

 zero current over the half of the line next to the generator; 

 15,000 volts and 0.917 ampere over the distant half of the line. 



11. The value of V L/C for an air line (two wire line) is 706 

 "ohms," and the value of V L/C for an underground cable is 

 loo "ohms." A ribbon wave in which E = 1,412 volts comes 

 over the air line to the point where the air line connects to the 

 cable, (a) What is the voltage and current in the wave which 

 enters the cable, and (b) what is the voltage and current in the 

 reflected wave? Ans. (a) + 349 volts and + 3.49 amperes, 

 (b) 1,063 v lts and + 1.51 amperes. 



12. A battery of which the resistance is 500 ohms and of which 

 the electromotive force is 8,000 volts is switched on to a line for 

 which the value of V L/ C is 1 ,000 " ohms. " Find the values of 

 voltage and current in the ribbon wave which shoots out on the 

 line. Ans. 5,333 volts, 5.333 amperes. 



Note. Let E and / be voltage and current in the wave. Then E/I = 1,000 

 and E + RI = 8,000, where R is the resistance of the battery. 



