AP EAPCET 24-Aug-2021 Shift 2 Paper

Let i1 and i2 be current drawn from battery 15 V and 30V, respectively as shown below Hence, by KCL.
Net current passing through branch BD
i3=i1+i2 ...(i)
By KVL, in loop ABDA,
15−6i1−3(it+i2)=0
⇒−6i1−3i1−3i2=−15
⇒3i1+i2=5 ...(ii)
In loop CBDC,30−3i2−3(i1+i2)=0
⇒i1+2i2=10 ...(iii)
By solving Eqs. (ii) and (iii), we gel
i1=0 and i2=5A
Substituting these values in Eq. (i), we get Current through branch BD,
i3=i1+i2=0+5=5A