Question

Consider the circuits shown below. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) ISR, = 6 kg HIER = 6 kN , 1 V = 2.4 v = V, = 1.9 V EV = 2.4 V Va = 1.9 V R, = 3 k2R , = 3 k22 R2 = 3 ks R2 = 3 k12 ' A T I BI (a) What is the current through each resistor in part A (in ma)? (Indicate the direction with the signs of your answers.) ma ma ma What is the current through each resistor in part B in ma)? (Indicate the direction with the signs of your answers.) ma ma ma (c) What is the power dissipated or consumed by each circuit (in mW)? circuit A circuit B mw mw (d) What is the power supplied to each circuit (in mW)? mw circuit A circuit B mw

Answer 1

a)From Kirchhoff's voltage law to the left loop IR +1,R-V = 0 60001, +300012 - 2.4 = 0.........1 From Kirchhoff's voltage law to the circuit V+1,R2 +1,R, -V2 = 0 30001, +30001; +2.4-1.9=0 300012 +30001, +0.5 = 0........2 From Junction rule 12 =1, +1;.-.....3 By solving the equations 11 = 0.353 mA 12 = 0.093 mA 1. =-0.26 ma b) From Kirchhoff's voltage law to the left loop IR +12R-V1=0 60001, +300012 - 2.4 = 0...-------....4 From Kirchhoff's voltage law to the circuit -V, +1,R+I RR-V = 0 30001, +30001, -2.4-1.9=0 30001, +30001, - 4.3=0... ....5 From Junction rule 12 =1, +17.......6 By solving the equations 1 = 0.0334 mA 12 = 0.734 mA 13 = 0.70 mA c)The power dissipated in the circuit A is, P=IR +1}R,+IR = (0.353x10(6000)+(0.093x10 (3000)+(-0.26×10 (3000) = 0.976X10% w =0.976mW The power dissipated in the circuit B is, P=1+R+IZR2+1ZR = (0.0334x 10-* )* (6000) + (0.734x10- )° (3000)+(0.70x10- )* (3000) = 3.09x10 W =3.09mW

d) The power supplied in circuit A is, P=1,V+112 = (0.093x10" (2.4)+(-0.26x10' (1-9) = 0.2708x10' w =0.271 mW The power supplied in circuit B is, P=1_V1 +13V2 = (0.734x10°) (2.4)+(0.70x10°)(19) = 3.09x10° w 33.09 mW