EE103hw5%20sol

EE103hw5%20sol - Homework 5 solutions 1 AB(120 80/80 100...

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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 ...
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