EE103hw5%20sol

EE103hw5%20sol - Homework 5 solutions 1 AB {

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

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

Unformatted text preview: Homework 5 solutions 1 AB { [(120+80)//80]+100+80}//200=(57.14+180)//200=237.14//200 = 108.49 CD ({[100+200+80]//80}+120)//80=[(380//80)+120]//80=(66.08+120)//80= 55.94 2 The equivalent resistance of the network is ( 20 + 10 + 20 ) //18 + 20 + 10 //16 + 10 + 18 // R {[50 //18] + 30} //16 + 28 // R ({ ( } ) ) [11.68 + 28] // R = 39.68 // R ({13.23 + 30} //16 ) + 28 // R ( 43.23 //16 ) + 28 // R 24V = 13.33 1.8 A 1 1 1 - = R = 20 13.33 39.68 R Req = 3 The equivalent resistance of the network is ({(10 + 10 + 10) //10 + 10} //10) + 10 = 16.36 V 8.18V = = 0.5 A Req 16.36 I = 4 The voltage V0 is V0 = R I = 10 K 0.005 A = 50 V The current in the dependent current source is 50 mA The current through the 20 K resistor is I 20 K = 50 mA 5 K = 10 mA 25 K 2 The power dissipated in the 20 K resistor is P20 K = I 2 R = (10 mA ) 5 20 K = 2 W 100V R The two extreme positions are 100V 1A = ( R + Rx ) 10 A = Solving for R and Rx 6 for the LOW position 5 A = for the MEDIUM position 10 A = and for the HIGH position 15 A = solving for R1 R2 and R3 R = 10 Rx = 90 6V 0.01 + 0.02 + R1 + R2 + R3 6V 0.01 + 0.02 + R2 + R3 6V 0.01 + 0.02 + R3 R1 = 0.6 R2 = 0.2 R3 = 0.37 7 The current through the device 1 is I = P 0.22W = = 10mA V 22V 2 V 2 ( 22V ) The equivalent resistance of the device 1 is Rdev1 = = = 2200 P 0.22W P 0.1W The current through the device 2 is I = = = 11mA 9V V V 2 ( 9V ) The equivalent resistance of the device 1 is Rdev1 = = = 810 P 0.1W 2 The circuit is equivalent to With these data Thus R2 = I1 = 40 mA I 2 = 29 mA V 9V = = 310.2 I 2 29 mA From the total current V0 = 24V - 2 50mA = 23.9V thus R1 = Finally R3 = V0 - V2 14.9V = = 372.5 I1 40 mA V0 - 22V 1.9V = = 190 10mA 10 mA ...
View Full Document

Ask a homework question - tutors are online