Question

get i with superposition theorem

V ŽUAZI 1210 12V o 6 kn 3 nš U19

Answer 1

#1 12 kn} 12k2 12V GMA 6K23 • 6k2 I I=2 using super position theorem . Sol: Super Position Theorem: In any linear, active, active, bilateral netwook consisting of number of energy sources, resistances etc..., the effect produced in any element when act element when act at a is equal to sum of effects produced by the same element when is taken individually While applying super position theorem, each sub circit has only one independent source. The other independent are de-energized. the time each source -> Sources ice, Voltage source → shoot citurit Current source + Open circuit Dependent sources cannot be supressed

case i) when 12 V Source acting alone, (om han open circuit + 1262 Zizka 12V (+ 12V 12 V I] 6Ko {6km l I ! , - ok 12 (12+) K 442 Isk 2 mA I = 2 mA Case (ii) when 6mA source acting alone > 12 V Source ) - short üsuit OV OV TA 12 KRZ 12kr 6mA 1 Iz 6kv2 6k2 OV Iz (reference ground)

Re-drawing the circuit 6mA Iz Zokn 12K2' Zizko Zokni'; Sparallel in connection parallel in connection RI GK2 || 12 kr I RT → resultant 2-(12x6) KXK equivalent resistance 3-4 (12+624 Ry=R, R2 42x6k 188 RitR₂ = 4kn 6mA V + 4kn v z 4 km Kve → Voltage across 4kiz resistor V=44k)(6m) V = 24 volts Note that: 4kV is the equivalent resistor of 6K & 12 kve resistor parallel, the voltage across each be equal. → AZ means voltage across 6k2 & 12kn resistor is 24 volts when resistors are will

IN I v 6k 244 I=4 mA ok By superposition theorem I Ist Iz 2 m .mt4.m 2+12 3 312) 4.6667 mA I - 14 mA 3