Here isa link to the circuit in question. Use Nodal analysis todetermine the cur
ID: 1829495 • Letter: H
Question
Here isa link to the circuit in question. Use Nodal analysis todetermine the current I1.
Here isa link to the circuit in question. Use Nodal analysis todetermine the current I1.
Explanation / Answer
(a) At node V1 Total current entering = 10mA Total current leaving = I3 + I2 +I1 Therefore at node V1 : I3 + I2 +I1 = 10mA --> (1) (b) At node V2 Entering Current =I2 + I3 Leaving current =I4 + 50mA Henceat V2 : I2 + I3 =I4 + 50mA ---> (2) (2) - (1) => I4 + 50mA -10mA =-I1 => I4 = -I1 - 40mA --> (3) (c) Using KVL in Loop of 1kohm , 2kohm , 2kohm resistors we get , (2k)*I2 + (2k)*I4 - (1k)*I1 = 0 => I1 = 2(I2 + I4) -----> (4) (d) Using KVL in Loop of 10k and 2k (where I2 and I3flows) we get , (10k)*I3 - (2k)*I2 = 0 => I2 = 5I3 -----> (5) Now put using (5) and (1) we get , I1 + I2 +5*I2 = 10mA => I2 = (10mA -I1 ) / 6 ---> (6) Using (3) , (4) and (6) we get I1 = 2[ (10mA - I1)/6 + (-I1 -40mA) ] => I1 = 20mA - I1/3 -2*I1 -80mA => (1+1/3+2)I1 = -60mA =>(10/3)I1 = -60mA => I1 = -60*3/10mA = -18mA Therefore I1 = -18mA (the negative sign indicatesthat actual current is opposite to the direction shown infigure)