Soln0533 080218YES - Hence v i = 4V. V 18 ) 4 ( 2 3 v o = =...

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Chapter 5, Solution 33. After transforming the current source, the current is as shown below: This is a noninverting amplifier. i i o v 2 3 v 2 1 1 v = + = Since the current entering the op amp is 0, the source resistor has a OV potential drop.
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Unformatted text preview: Hence v i = 4V. V 18 ) 4 ( 2 3 v o = = Power dissipated by the 3k resistor is = = k 3 324 R v 2 o 108 mW =-=-= k 1 18 12 R v v i o a x 6mA 4 k 2 k 12 V +-+-1 k v o v a v i 3 k...
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