Question

Find current in the branch CD of the circuit

Answer 1

Therefore $R_{eq}=\frac{3}{2}\Omega$
from ohm's law, current supply by cell
$I=\frac{V}{R_{eq}}=20A$

using KCL, current division is shown below
(current gets equally divided in parallel branches with same resistance)


using KCL at junction C
Current in branch CD
$I_{CD}=I_{AC}+I_{BC}$
$I_{CD}=10\Omega +5\Omega =15A$

Hence current in CD = 15 A
Hence, option (a) is correct