TS EAMCET 04 May 2019 Shift 2 Solved Paper

Consider the circuit diagram shown below.
Apply KVL in loop (1).
−5+10I1+&10(I1−I2)=0
20I1−10I2‌‌=5‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)
Apply KVL in loop (2)
10I2+3&+10(I2−I1)=0
−10I1+20I2=−3
−20I1+40I2=−6‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II)
Solve equation (I) and (II).
I2=−‌

1

30

A
And,
I1=‌

7

30

A
Current in branch AB is,
IAB‌‌=I1−I2
‌‌=‌

7

30

−(−‌

1

30

)
‌‌=‌

8

30

A
Therefore, the voltage across AB is,
VAB‌‌=IABRAB
‌‌=(‌

8

30

A)(10Ω)
‌‌=‌

8

3

V