In the circuit shown in fig. find the current through the branch BD.

Options

A. 2 A
B. 3 A
C. 4 A
D. 5 A

Answer

D

Explanation

Method 1 : Using Kirchhoff's law M The currents in the circuit are assumed as shwon in fing Applying KVL along the loop ABDA, we get
`-6i_1 - 3i_2 +15 = 0`
or `2i_1 + i_2 = 5 (i)`
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`.

Method 2: using equivalnet battery method (i)
we can treat `6Omega and 3 Omega` resistance connected with 15V
and 30V batteries, respectively, as internal resistances of the
batteries. Now we can assume these batteries are connected in
parallel, and equivalnet battery of these two is connected with
resistance connected across B and D.


Hence, current through resistace connected across B and
D is
`i = (20)/(2+3) = 5A`
Method 3: Using equivalent battery method (ii)
We cna assume a battery of zero emf is connected in series
with resistance connected across BD. This resistance can be
treated as internal resistance of zero emf battery. Now we have
three battery which is the open cirucuit.


As equivalent battery in as open circuit, the potential
difference across each of the battery branch wikll be 15V.
Hence, current through the resistacne connected across BD is
`I = (150)/(3) = 5A`
Method 4 : Using nodal method
Let us asume the potential of node D zero and the potential of node B x. then we can assign the potentials at different nodes
as shown in fig 5.124. At node B,
Hence current through resistacne connected across BD is
`i = (15 -0)/(3) = 5A ("from" B to D)`

Method 5: using superposition principle In superposition method, we can take the effect of one battery at a time. From fig, it is clear that the effect on two
batteries can be taken by adding the effect of individual batteries.