Unformatted text preview: ed by the resistor. Thus: Ps 72W Alternatively: notice the signs (active sign convention)
so this is provied power Ps VI (6V )(12 A) 72W 15.10 Examples The power absorbed by the resistors are known to be: P1 20W P2 40W P3 60W (i) Find the power supplied by the source. Ans: Due to conservation of power the power delivered by the source equals the power absorbed by all resistors. Thus: Ps 20 40 60 120W ii) Compute the current I Ans: From Ps VsI (notice that the signs are correct based on the fact that this is delivered power) we get I Ps Vs 120 60 2 A iii) Compute the voltage drop across each resistor. Ans: Notice that since each element is a 2‐terminal circuit element, the current entering equals the current leaving. i.e. I1 I2 I3 I4 P
Hence from V1 1 we get I 15.11 V1 20 10V 2 Similarly P 40 20V
V2 2 I
2 P3 60
V3 30V
I
2 Now look at something interesting: 10V + 20V + 30V = Vs ! This is a concept we will learn next time. (iv) Find the resistance of each resistor: Ans: From Ohm’s law R V I V 10
Thus R1 1 5 I
2 V 20
R2 2 10 I
2 V 30
R3 3 15 I
2 Kirchhoff’s Laws (we have seen them, this is the specifics of the formulation) Kirchhoff’s laws are used to analyze a circuit (i.e. solve it – find everything you need/want). There are two of them: Kirchhoff’s current law (KCL) Kirchhoff’s voltage law (KVL) In order to understand these laws we need a few definitions: Terminology ( = Kirchhoff’s language) Series connection of two‐terminal lumped elements = = sequential connection, end‐to‐end connections 15....
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This note was uploaded on 03/06/2014 for the course ECE 201 taught by Professor All during the Spring '08 term at Purdue.
 Spring '08
 ALL
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