Question

Current I in the network shown is:
1) 2A
2) 0.5A
3) 2A
4) 5A​

Answer 1

OptionA is the correct answer

Answer 2

Explanation:

Here, current I is divided into two

I=I1+I2.....(1)

In upper loop, applying Kirchhoff's law-

(going clockwise)

0=10*I1+6*I1-16*I2

0=16*I1-16I2

0=16(I1-I2)

=> I1-I2=0

=> I1=I2......(2)

In lower loop, applying Kirchhoff's law-

(going clockwise)

10=16*I2+2I

10=16*I2+2(I1+I2)

10=18*I2+2I1

10=20*I1 (using (2))

I1=0.5A=I2

hence,I=I1+I2= 0.5+0.5A=1A

(kindly recheck your options)

Hope it helps!