Question

In the given circuit diagram, a wire is joining points $B$ and $D$. The current in this wire is

• A. $Zero$
• B. $0.4A$
• C. $4A$
• D. $2A$

Answer 1

The correct option is D $2A$

Let us first find the equivalent resistance across cell.

Reducing the circuit by solving parallel and series combination of resistances, we have

Parallel combination of $1\Omega$ and $4\Omega$

$=\frac{1\times 4}{1+4}=\frac{4}{5}\Omega$

Parallel combination of $2\Omega$ and $3\Omega$

$=\frac{2\times 3}{2+3}=\frac{6}{5}\Omega$

Series combination of $\frac{4}{5}\Omega$ and $\frac{6}{5}\Omega$

$=\frac{4}{5}\Omega +\frac{6}{5}\Omega =2\Omega$

From the circuit diagram ,
it is clear that

$I=\frac{20V}{2\Omega }=10A$
(ohm’s law )

Voltage drop across $\frac{4}{5}$ $\Omega =I\times \frac{4}{5}$

$=10A\times \frac{4}{5}$ $\Omega =8V$
(ohm’s law )

this is voltage drop between $A$ and $B$ (or $D$)
Voltage drop across $\frac{6}{5}\Omega$

$=I\times \frac{6}{5}=10A\times \frac{6}{5}\Omega =12V$

(ohm’s law )
this is voltage drop between $B$( or D) and $C$

voltage drop between $A$ and $B$ (or $D$) is $8V$ and voltage drop between $B$(or $D$) and $C$ is $12V$

Referring to circuit diagram
from ohm’s law
Current flowing through $4\Omega$

$=\frac{8V}{4\Omega }=2A$

Current flowing through $1\Omega$

$=\frac{8V}{1\Omega }=8A$

Current flowing through $2\Omega$

$=\frac{12V}{2\Omega }=6A$

Current flowing through $3\Omega$

$=\frac{12V}{3\Omega }=4A$

$\therefore$ By applying $KCL$ law at junction $B$ ,
current in branch $BD=8A-6A=2A$

(total incoming current $=$ total outgoing current )

Thus, current flowing in $BD$ is $2A$

Hence option (B) is correct.