Question

IV. Experiment Consider the circuit diagram below: 1 ko 3.6 ko 4.7 kΩ 10 V 2 3 ΚΩ 3 kΩ Write the node equations at node 1 and 2 Solve the node equations for the node voltages V, and V,

Answer 1

we have to apply the Kirchoff's Current Law (KCL) at node1 and node2 and calculate the expression of voltages V1 and V2 in terms of I , then substituting the value of in terms of V1 and V2 calculate the values of V1 and V2 from the formed equations. Please find the attached solution.

А Sall 1 kng 3.6kr lov 4.7kR I 3 kr mm & k2 B using Kirchoff's Current Law (KCL) at Nodel :- V - VA 1 V - Ve 3 Vq - ķ + + O = 4.7 V-10 1 + V-o I =0 + 3 V.- 10 + Vi 3 + =0 4 V t I = 10 V, 10-I 4 40 { NOTE: In above circuit as all the resistances valves hare given in (KR), so the current values will be in (mA) (MA) ie in (mA). I is

using Kirchoff's Current Law (KCL) (P.2 at node 2 :- W-VA - VR + + - V, 4.7 = 3.6 3 Va-10 3.6 + V₂ - 0 ㅋ (-I) = 0 + 3 10 3.6 + - I - V2 3 =0 3.6 3 V2 10 (3) + 3.6 V₂ 10.8 10.8 I 个 - O → 6.6 V₂ - 30 – 10.8 I = → 6.6 V₂ - 30- 10.8 I 30 6.6 + 10.8 6.6 I 9.

(P3 (ii) solve To Node voltages y, and V2: using egno:- V 10- 4 {since, I = ~, -V₂1 4.7 10 10 VI-V2 (4.7) V = 10 4 Vi 18.8 + 188 → [ 1 + 11 18.8 V2 18.8 + 10 ا 19.8 V V2 18.8 10 F → 1.05319 0.05319 V₂ +2.5 1.05319 V 0.05319 V = 9.5 9.

4 using esa : I V₂ - 30 6.6 + 10.8 6.6 + 4.5 45 46 = 18 () V₂ = 4.54546+ 0.34816 v 0.34816 V4 » 1.34816 V2 - 0.34816 = 4.54546 4) Multiplying est by (1-34816) and es by (0.05319) and Adding both equs :- -1.34 816) (0.05319)V2 + (1.34816)(1.05319)V) =(1-3481672:s) + (1-34816) (0.05319) va 6.34816)(0.05319)V, = (0.0531974.sys 54546 1.40135 V 3.6121727 V 2.5776 Volt Substituting valve of 'vi in eq" V = 4.03727 Volt