Question

For the circuit shown in the figure. Find out the current in the wire $BD$.

• A. $2\text{A}$ from $D$ to $B$
• B. $1\text{A}$ from $D$ to $B$
• C. $3\text{A}$ from $D$ to $B$
• D. $4\text{A}$ from $D$ to $B$

Answer 1

The correct option is B $1\text{A}$ from $D$ to $B$

Step $1$: Assume some current in branches taking $\text{KCL}$ in to account at the junction


Step $2$: Apply $\text{KVL}$ in closed loops,

In loop $DCBD$,

$\Rightarrow 20-5i=0$

$i=4\text{A}...\left(1\right)$

In loop $BADB$,

$+10-2(i-i_1)=0...\left(2\right)$

Using Eq.$\left(1\right)$ and $\left(2\right)$,

$10=2(4-i_1)$

$i_1=-1\text{A}$

The diagram showing final distribution of current in branches


The negative sign of current indicates that its direction will be in the direction opposite to that assumed initially.

$\therefore 1\text{A}$ current flows from $D$ to $B$

Method $2$

Step $1$ : Lets assume potential at point $B$ to be zero. Now the points $B$ and $D$ are connected by same conducting wire, hence both will have same potential i.e $0\text{V}$

Step $2$ : Mark the potential at other points


Step $3$ : Find the current through each resistance by applying ohm's law

$i=\frac{\text{Potential difference}}{\text{Resistance}}$

$\Rightarrow i_{CB}=\frac{20-0}{5}=4\text{A}$ and $i_{AD}=\frac{10-0}{2}=5\text{A}$

Step $4$ : Apply $\text{KCL}$ and obtain required currents in the branches.


Thus in branch $BD$, $1\text{A}$ current flows from $D$ to $B$

Why this question? Tip: In such problems assigning $\text{'Zero'}$ potential at the nodes is recommended to find the current in respective branches, as it will save a lot of time in competitive examinations.