Question

In the given circuit diagram, a wire is joining points $B\&C$. Find the current in this wire

• A. $0.4A$
• B. $2A$
• C. $0$
• D. $4A$

Answer 1

The correct option is B

$2A$

Solution:

Step 1: Given data

Observe that from the figure, $1\Omega , 4\Omega \& 2\Omega , 3\Omega$ in parallel connection separately, and $1\Omega \parallel 4\Omega$ is in series with $2\Omega \parallel 3\Omega$.

Step 2: Calculate the $R_{eq}$

Knows if $R_1\&R_2$ are in parallel then $R_{eq}=\frac{R_1\times R_2}{R_1+R_2}$, and if $R_1\&R_2$ are in series then $R_{eq}=R_1+R_2$.

Consider the required formula, $R_{eq}=\frac{R_1\times R_2}{R_1+R_2}+\frac{R_3\times R_4}{R_3+R_4}$, where $R_1,R_2,R_3,\&R_4$ are $1\Omega , 4\Omega , 2\Omega ,\& 3\Omega$, respectively.

Therefore,

$\begin{array}{l}\Rightarrow R_{eq}=\left(\frac{1\times 4}{4+1}\right)+\left(\frac{2\times 3}{2+3}\right)\\ \Rightarrow R_{eq}=\frac{4}{5}+\frac{6}{5}\\ \Rightarrow R_{eq}=\frac{10}{5}\\ \Rightarrow R_{eq}=2\Omega \end{array}$

Step 3: Calculate the current flow

Knows $i=\frac{V}{R}$

where $i=$current, $V=$voltage, and $R=$Resistance.

Therefore,

$$\begin{gathered} \Rightarrow i=\frac{20}{2} \\ \Rightarrow i=10A \end{gathered}$$

Step 4: Draw the diagram

Know that the current will distribute in a ratio opposite to resistance.

Therefore, distribution will be as

Step 5: Calculate the current through $B\&C$

Now, from the diagram, the current in branch $BC$:

$\begin{array}{l}\Rightarrow I_{BC}=\left(\frac{4}{5}\times 10\right)-\left(\frac{3}{5}\times 10\right)\\ \Rightarrow I_{BC}=8-6\\ \Rightarrow I_{BC}=2 A\end{array}$

Hence, the correct answer is option (B).