20105ee115A_1_hw2_solns

20105ee115A_1_hw2_solns - 1) In the circuit shown below,...

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1) In the circuit shown below, find the voltage on node x. Let Vx be the voltage at node X. Let the diodes be labelled as below: We make the following observations from this circuit : a) Vx has to be greater than 1 V as the current has to flow to the right to support this Vx. b) 1<Vx<2 : Diodes D3 and D4 are now reverse biased and the circuit reduces to :
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Applying Kirchoff Current Law to find the voltage at node X : 6 − ܸݔ ܴ + 2 − ܸݔ ܴ = ܸݔ − 1 ܴ Therefore Vx = 3V. But Vx > 2 V. This is contradicting our assumptiom in the beginning. Hence this voltage is not valid c) Vx > 2V D1 , D4 and D5 are forward biased. Applying Kirchoff Current Law to find Vx: 6 − ܸݔ ܴ = ܸݔ − 2 ܴ + ܸݔ − 1 ܴ Therefore Vx = 3 V ANSWER : 3V 2) In this circuit, all the capacitors are the same size (C) and the voltage written by each of them is the initial voltage on them. Find the final value of the voltage across C 4 .
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Let V1, V2, V3 and V4 be the voltages across C1, C2, C3 and C4 respectively. A) The diode D1 only allows C2 to charge up. Hence V2 increases. B) D2 only allows C3 to discharge. Hence V3 cannot increase. C) C3 cannot discharge through C2 either, since D2 is always OFF even if D1 is forward biased Hence the equivalent circuit is as follows :
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This note was uploaded on 01/16/2011 for the course EL ENGR 115A taught by Professor Abidi during the Fall '10 term at UCLA.

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20105ee115A_1_hw2_solns - 1) In the circuit shown below,...

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