Question

In the given circuit, the value of current (in $\text{mA}$) in the wire between points $A$ and $B$ will be

• A. $8\text{mA}$
• B. $12\text{mA}$
• C. $16\text{mA}$
• D. $4\text{mA}$

Answer 1

The correct option is A $8\text{mA}$


On observing the symmetry of resistance connected in circuit, we can infer that the current in $1$ $\text{k}\Omega$ and $2$ $\text{k}\Omega$ will be equal respectively.

As shown in figure, $i_1$ & $i_2$ current passes through them.

$\Rightarrow i_{AB}=i_1-i_2$

The equivalent circuit can redrawn as:


$i=\frac{V}{R_{eq}}$

$\Rightarrow i=\frac{32}{[\left(\frac{2\times 1}{2+1}\right)+\left(\frac{2\times 1}{2+1}\right)]}=\frac{32}{\left(4/3\right)}$

$i=24\text{mA}$

From the figure,

$V_1+V_1=32$

$V_1=16\text{V}$

Now current through $1$ $\text{k}\Omega$ and $2$ $\text{k}\Omega$ resistors will be,

$i_1=\frac{V_1}{R_1}=\frac{16}{1\times {10}^3}=16\times {10}^{-3}$

$\Rightarrow i_1=16\text{mA}$

Similarly, $i_2=\frac{V_1}{R_2}=\frac{16}{2\times {10}^3}=8\text{mA}$

Hence current in wire between $A$ & $B$,

$i_{AB}=i_1-i_2=16-8=8\text{mA}$.

Why this question? It intends to test your understanding of ''role of symmetric arrangement of resistors in current distribution in a circuit''.