Question

Problem 6 Using Kirchhoff's rules, find the following. (& = 71.0 V, & = 60.2 V, and E3 = 79.2 V.) 4.00 k12 b d с 82 2 Rg 8 R2 3.00 k12 2.00 k2 w R (a) the current in each resistor shown in the figure above (b) the potential difference between points c and f (c) which point, corf, is at the higher potential?

Answer 1

SOLUTION 4K2 d I I 2 Iz R3 Ez a Iz E I & 3 R₂ & 3 kr 2KR YIz I3 R, If By Kirchhoff's current law:- I t I3 = I2 By kirchhoff's voltage law:- Take : voltage gain ove Voltage drop ove For loop - E, E2 - I 2 R2 - I, Ri=0 IIRI + I2 R2 = E, - E2 II (2x103) + Iz (3x103) = 71-60.2 - 2 I + 3 I 2 10.8 1000 Put value of I2 = I t Iz in above equation = > 2 I, + 3(I +F3) = 10.8 1ooo = ) 5I1 + 3 I3 = 10.8 looo - <i>

from loop :- Ez - I3 R3 - E, - I2 R2 = 0 > I3 R3 + I2 R₂ = € 3-E2 Is (4x103) + I2 (3x103) = 19 4 I 3 + 3 I 2 = 19 103 Put value of Iz = I + Iz in above equation 4 I3 + 3 (II + I3) - 19 10oo 7 I3 + 3 I 1 = 19 1000 Multiply equation (ii) with 3 15 I, + 9 I3 = 32.4 lie lood Multiply 35 I3 + 1511 = equation <iii) with 5 le 95 looo - (v) Sub stract equation civ) from (u) → 26 I 3 = 62.4 1ooo Iz = 2.4077 2.4 mA

Put I 3 in equation Lily 51, + 3x24077 1 000 10.8 1ooo IL 51 = 10.8 1ooo 7.223) looo 3.5769 looo i. Il = 0.71538 looo 0.72 MA Now, I 2 = It I3 = 2.4+0.72 Iz = 3.12 MA art Lay current in R, = I, = 0.72 m A current in R2 = Iz 3.12 MA current in R3 = 13 2.4mA Part (6) Vc - E - Iz (R2) =Vf i. Vc - Uf = Ezt Iz R2 = 60-2 + (3•12X163) (3x103) - Goo2+ 9.36 69.56 Part (C) from part (b) we can say point c is at higher potential.