Question

18 V W HA w 1) Current cach branch. 2) Power Dissipated by cach resistor. 3) Power dissipated by cach source. 4) Potential difference between points (a) and (b).

Answer 1

left lov sr Right bar a futia 12V Hamel 702 Ik 1 3 fis 5 361 from Kirchoff's law in upper loop 12 – 7i2 - 5 x (1 +i2) +18 - 8x (i1 +i2)-11i2 = 0 30 – 13 X (i1 + 2) - 1812 = 0 13i1 +31i2 = 30 from Kirchoff's law in lower loop 36 + 7i2 - 12 +11,2 – 5ij = 0 24 + 1812 – 5ij = 0 511 - 1812 = 24 solving the equations

i = 3.3 12 = -0.416 Current in upper branch 11 + i = 3.3 -0.416 = 2.884A right to left Current in middle branch i2 = 0.416A right to left Current in lower branch i1 = 3.3A left to right Power dissipated by upper 5 ohm P = (ij +iz) x (5) Ps = (2.884)2 x (5) = 41.587W Power dissipated by lower 5 ohm Pa = (11)? (5) Ps = (3.3)2 x (5) = 54.45W Similarly Pg = (i + i2)2 x 8 Pg = (2.884) (8) = 66.54W Pui = (iz)? *(11)=(-0.416)2 x (11) = 1.904W

Pa = (12)2 x (7) = (-0.416)2 x (7) = 1.211W Power dissipated by 18 V P18 = {(i1 + i2) = (18) (2.884) = 51.912W Power dissipated by 12V P18 = 12i2 = (12) Ⓡ (0.416) = 4.992W Power dissipated by 18 V P18 = E(i1 + i2) = (18) Ⓡ (2.884) = 51.912W Power dissipated by 36 V P36 = 36i1 = (36) X (3.3) = 118.8W Now Potential difference between a and b Va - 5i1 + 36 = V V -V= 5ij – 36 = 5 x (3.3) – 36 = – 19.5V