In the given circuit diagram, the currents, I1 = -0.3 A, I4 = 0.8 A and I5 = 0.4 A, areflowing as shown. The currents I2 , I3 and I6 ,respectively, are :

Answers (1)

 

Kirchoff's first law -

\sum i=0

i_{1}+i_{3}=i_{2}+i_{4}

- wherein

 

Apply Kcl at Q

I6+I3 = I1+ I2   .........(1)

I1 = 0.3A

I4 = 0.8A

I5 = 0.4A

KCL at 5

I4= I5 +I3

I3 = I4 - I5

    = 0.8 +0.4

    =0.4A

KCL at R

I1+I2 = I4

I2 = I4 - I1

    = 0.8 - (-0.3)

    =1.1A

From equation (1)

I6 + I3 = I1+ I2

I6 = I1 + I2 - I3

    = -0.3+1.1 -0.4

    =0.4A

I2 = 1.1A

I3 = 0.4A

I6 = 0.4A

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