SingleLoop7Ed

# 30 10 30 10 3 2 4 2 ω ω resistor 30k on power the

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30 ) 10 * 30 ( ) 10 ( 3 2 4 2 = - = = - RESISTOR 30k   ON   POWER

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THE “INVERSE” VOLTAGE DIVIDER = + = k V S 500 3 . 458 220 20 220 DIVIDER    INVERSE" "     + - 1 R 2 R S V - + O V S O V R R R V 2 1 2 + = V O L T A G E   D I V I D E R O S V R R R V 2 2 1 + = " I N V E R S E "   D I V I D E R S V    COMPUTE APPLY KVL TO THIS LOOP mA I kI kI 05 . 0 0 40 12 80 6 - = = + + + - V V V kI V bd bd 10 0 12 40 = = - - = - + V 3 V V S 9 3 20 20 15 25 = + + = INVERSE DIVIDER PROBLEM
+ - + - x V 3 - + V 4 k 4 - + X V a b S V V V S 12 = source   dependent     the by    supplied or    absorbed power     the   is    ) 3 ( Vx P = = = ) 3 ( Vx ab x P V V EXAMPLE APPLY KVL TO THIS LOOP = + + + - 0 3 4 12 X X V V     : KVL V V X 2 = = + + 0 3 4 X ab V V    : KVL V V ab 10 - = 0 = - + X S ab V V V    : KVL I ) CONVENTION   SIGN   (PASSIVE    I V P X V X 3 ) 3 ( = mA k V I 1 4 4 = =    : LAW   OHMS' mW mA V P X V 2 ] [ 1 * ] [ 2 ) 3 ( = = + - + - k 30 V 9 k 20 k 10 = = = DE CD DA I V V A B C D E EXAMPLE DETERMINE I USING KVL I 0 I * 10k I * 30k 9 I * 20k 12 -    : KVL = + + + + mA k V I 05 . 0 60 3 = = mA 05 . 0 V I k 5 . 1 * 30 = DA  V FOR KVL  0 * 10 12 = - + I k V DA    : KVL V V DA 5 . 11 - = + - ab V KVL HERE OR KVL HERE
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• Spring '12
• HaroldKlee

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