Question

The value of current $i_1$ flowing from $A$ to $C$ in the circuit diagram is :

• A. $4A$
• B. $5A$
• C. $2A$
• D. $1A$

Answer 1

The correct option is D

$1A$

Explanation for correct option

Step 1: Find the equivalent resistance

From the above circuit diagram, it is clear that the circuit is forming a balanced Wheatstone bridge

When the current flow from $A$ to $B$ and $A$ to $D$ the potential at $B$ and $D$ is same.

So no current pass through $B$ to $D$

Now the circuit diagram is

Here R1 and R2 are in series

Similarly, R3 and R4 are in series

and, R5 and R6 are in series

Let $R_{eqv}$ is the equivalent resistance.

$$\begin{gathered} \therefore \frac{1}{R_{eqv}}=\frac{1}{2+2}+\frac{1}{4+4}+\frac{1}{2+2} \\ \Rightarrow \frac{1}{R_{eqv}}=\frac{5}{8} \\ \Rightarrow R_{eqv}=\frac{8}{5} \end{gathered}$$

Step 2: Find the current $I$

Let the current be $I$

We know that $V=IR$ where $V$ is the voltage , $I$ is the current, $R$ is the resistance.

Given $V=8V$

$$\begin{gathered} \therefore I=\frac{V}{R} \\ =\frac{8}{{\displaystyle \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}} \\ =5A \end{gathered}$$

Step 3: Find the value of $I_1$

From Kirchhoff's circuit laws we have

$I=I_1+I_2+I_3$

The equivalent resistance from where current $I_2+I_3$ passes is

$$\begin{gathered} \Rightarrow \frac{1}{R_{eqv}}=\frac{1}{2+2}+\frac{1}{2+2} \\ \Rightarrow \frac{1}{R_{eqv}}=\frac{1}{2} \\ \Rightarrow R_{eqv}=2 \end{gathered}$$

The resistance $4\Omega$ is in series connection .

So the equivalent resistance is $8\Omega$

We know that the current is inversely proportional to the resistance

Let $I_2^{\text{'}}=I_2+I_3$, $R_1=8,R_2=2$

$$\begin{gathered} \therefore \frac{I_1}{I_2^{\text{'}}}=\frac{R_2^{\text{'}}}{R_1} \\ \Rightarrow \frac{I_1}{5-I_1}=\frac{2}{8}\left[\because I_1+I_2^{\text{'}}=5\right] \\ \Rightarrow 4I_1=5-I_1 \\ \Rightarrow 5I_1=5 \\ \Rightarrow I_1=1A \end{gathered}$$

Hence option D is correct