HW3solns

# HW3solns - !"#\$%&'()*5( Find I0 in the circuit...

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Unformatted text preview: !"#\$%&'()*5( Find I0 in the circuit using nodal analysis 3k 9V 2k 6k 2k +,--&./&0(+#%,/1#2( V1 3k 9V V2 2k V3 6k 2k V1 = 9V 9 - V2 3k = V2 V2 - V3 + 6k 2k V3 = 1.2V I 0 = 0.6mA V3 - V2 V3 V3 1.2 + + = 0 V3 = 1.2V , I 0 = = 0.6mA 2k 2 k 2k 2k !"#\$%&'(!"#\$( Use nodal analysis to find out I0 and I1 6k 12mA 4k I0 3k 12V 2k +,--&./&0(+#%,/1#2( V1 6k V2 3k 12V 4k I0 2k 12mA I1 V1 = -12V ; I1 = -6mA 12 V1 V1 - (-12) + + = 0 V2 = -6V 1K 4k 2k V1 = 8V ; I 0 = 2mA V1 = 8V ; I 0 = 2mA !"#\$%&'().24( Find I0 in the network shown. 1k 12V 1k 1k 1k I0 6mA +,--&./&0(+#%,/1#2( 1k V2 V1 1k 1k 1k I0 6mA 12V V2 -12 V1 V1 - V2 V1 - (V2 - 12) + + =0 1k 1k 1k V2 - V1 V2 V2 - 12 - V1 6 + + + =0 k 1k 1k 1k 3V1 2V2 -12 -2V1 3V2 6 - = + = ; 1k 1k 1k 1k 1k k 3V1 - 2V2 = -12 -2V1 + 3V2 = 6 -24 V1 3 -2 -12 5 V = -2 3 6 = -6 2 5 -24 V1 I 0 = = 5 = -4.8mA 1k 1k !"#\$%&'()*3)( Find Io in the network shown using loop analysis. Then solve the problem using matlab and compare your answers. I 0 = -4.8mA 6K 6K 6K 6K 6V 5mA IO +,--&./&0(+#%,/1#2( I3 6K 6K 6K I1 6V 5mA I2 6K IO 6 + 6 K ( I1 - I 3 ) + +6 K ( I 2 - I 3 ) + 6 KI 2 = 0 6 KI 3 + 6 K ( I 3 - I 2 ) + 6 K ( I 3 - I1 ) = 0 I 2 - I1 = 6K -6 K -1K 5 K 12 K -6 K 1K -12 K I1 -6 18 K I 2 = 0 0 I 3 5 Using Matlab EDU Z=[6000 12000 -12000;-6000 -6000 18000;-1000 1000 0] Z= 6000 -6000 -1000 12000 -6000 1000 -12000 18000 0 EDU V=[-6;0;5] V= -6 0 5 EDU I=inv(Z)*V I= -0.0046 0.0004 -0.0014 !"#\$%&'()*70( Find Vo in the circuit shown. 5K 1K + 1V VO - +,--&./&0(+#%,/1#2( KCL at the intersecting input 1-0 0 -V 0 = 0.1K 5K V 0 == -5V !"#\$%&'()*73( V 0 == -5V The network shown is a current-to-voltage converter or transresistance amplifier. Find Vo/Is for this network. 1 is VO - +,--&./&0(+#%,/1#2( KCL at the inverting input is 5 - 4 0 - V0 = 1 1 V0 = -1 is 0 V0 = -1 is 0 ...
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## This note was uploaded on 10/12/2009 for the course ECE E 245B taught by Professor Uftureli during the Fall '01 term at Stevens.

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