In the circuit shown, find the current through the branch BD.

Explanation

To find the current in a particular brance in a circuit
where resistances and voltages are given, the lower loop rule
is applicable. In orber to find out the solution, the given circuit
is directed into parts or closed loops, which are analyzed using
Kirchhogg's second law and corresponding sign convantions.
Assume the currents in the ciruit as shown in
applying KVL along the loop ABDA and moving in the
clockwise direction, we get
`-6I_1 - 3I_2 + 15 = 0`
or `2I_1 + I_2 = 5` (i)
While moving in the direction of the current corresponding
IR, products are taken as negative and if the negative terminal
of the battery comes first, then emf is taken as positive
Applying KVL along the loop BCDB, we get
`-3(I_1 - I_2) - 30 + 3I_2 = 0`
or `I_1 +2I_2 = 10` (ii)
Solving Eqs. (i) and (ii) for `I_2` we get` I_2 = 5A`