Calculate the value of current I1 , I2 , and I in the given figure , if E1 = 2V , E2 = 1V ; r1 = 2 ohm , r2 = 2ohm and R = 6 ohm
​​

Dear Student,
Applying kirchoff's second law to loop BCGFB we get
I2r2-e2+e1-I1r1=0------(1)
Applying kirchoff's second law to loop BCNMB we get
I2r2-e2+(I1+I2)R=0
I1R+I2(R+r2) = e2-------(2)
Applying kirchoff's second law to loop FGNMF we get
I1r1-e1+(r1+r2)R=0
I1(R+r1)+I2R=e1--------(3)
Solving (1) and (3) we get I1  and solving (1) and (2) we get I2 as follows
I1 =e1-e2R+e1r2Rr1+r2+r1r2I1 =2-16+2×262+2+2×2=514AI2 =-e1-e2R+e2r1Rr1+r2+r1r2I2 =-2-16+1×262+2+2×2=-114ASo I =I1+I2I =514+-114=414=27A
Regards.

 

  • -2
What are you looking for?