Electric Circuits 8th Edition 108

Electric Circuits 8th Edition 108 - the 5 kO resistor,...

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84 Simpte Resjrtiv€ Cjrcuits 3.20 a) Show that the current in the ,tth branch of the 6BG circuit in Fig. P3.20(a) is equal to tle source current is times the conductance of the *th branch divided by the sum of the conductanceq that is, iFr b) Use the result derived in (a) to calculate the cur- rent in the 6.25 f,) resistor in t]le circuit in Fig.P3.20(b). Figuru P3.20 Specify tle iesistorc ir the circuit in Fig. P3.21 meet the following design criteda: is=5mA;0s:lvti=4i21 i2 : 84; and i3 : 5la. Figue P3.21 Look at the circuit in Fig.P3.1(a). a) Use current division to fhd the current flowing from top to bottom ir the 10 kO resistor. b) Using your result from (a), find the voltage drop affoss the 10 kO resistor, positive at the top. c) Using your result ftom (b), use voltage diusron to find the voltage drop across the 6 kf,) iesistor, positive at t])e top. d) Using 'our resuit lrom (c). use
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Unformatted text preview: the 5 kO resistor, positive at the left. Look at the circuit in Fig. P3.1(b). a) Use voltage division to find the voltage drop across the 240 O resistor, positive at tle left. b) Using your result from (a), find tle current flow-iog i0 the 240 O resisror lrom lei lo right. c) Using your result form (b), use curent division to find the current in the 140 O resistor. 324 a) Find the voltage ?J, in the circuit in Fig. P3.24. """ b) Replace the 30V source with a general voltage source equal to 14. Assume % is positive at the upper teminal Find o, as a function of %. hg,rrc P3.24 3.2S Find ?1 and ?J2 in the circuit itr Fig. P3.25-EPr'r Figure P3,25 5ko 1ko 1 2 0 30v 60 ko 15 k{) 50() 321 3.26 Find o, in the cfcuit in Fig. P3.26. '5"!' tigure P3.26 10 ko 2 k O + i r , - 4 k O 15 kO 3ko 12 kO 327 Find i. and is in tlle circuit in Fig. P3.27. """ Figure tl,zz R. i. 6'75 V 10()...
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This note was uploaded on 07/05/2010 for the course EE 100 taught by Professor Boser during the Spring '07 term at University of California, Berkeley.

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