20081ee10_1_ee10_hw5_sol

# 20081ee10_1_ee10_hw5_sol - P 5.4 Since the current into the...

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Unformatted text preview: P 5.4 Since the current into the inverting input terminal of an ideal op—amp is zero, 0, therefore '00 = 9'01, - 8va the voltage across the 3.3 M9 resistor is (3.3 X 106)(2.5 X 104) or 8.25 V. Therefore the voltmeter reads 8.25 V. 'UbIOV, va=‘—12V P55 [a] ia2l22X10—6:20#A vb = 0 V, '00 -—18 V (sat) ’Ub"'2 V v ——10 V va=e20x103ia=—400mv _ 7 o_ va '03, 21a —— 0,, ”b = 2 V, 00 = ‘14 V ['3] 60,000 + 20,000 + 240,000 = 0 vb = 8 V, 120 = 18 V (sat) '. 0,, = 17m, = —6.8 V '00 = 40.5 — 80a 2 ilS [c] ia = 20,11A 2.8125 g va 3 7.31 ,5 V [.1] i0 = “’0 + ”a “”0 2 111.6714 A 80,000 240,000 P 5.28 up = on = mg Rbib -" 1000ia Rb’ib - ’00 1000 + Rf " 0 (Rb +53%: z—& 1000 Rf b a““Rf ’U _. [1%be °" 1000 + Rb] z'b —- sz'a Rf z 4000 (2 4R1) + Rh 2 4000 Rh 2 800 Q ...
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