Question

The figure shows a circuit having eight resistances of $1\Omega$ each, labelled $R_1$ to $R_8$, and two ideal batteries with voltages ${\epsilon }_1=12\text{V}$ and ${\epsilon }_2=6\text{V}$.



Which of the following statement(s) is(are) correct?

• A. The magnitude of current flowing through $R_1$ is $7.2\text{A}$.
• B. The magnitude of current flowing through $R_4$ is $2.4\text{A}$
• C. The magnitude of current flowing through $R_3$ is $4.8\text{A}$.
• D. The magnitude of current flowing through $R_2$ is $1.2\text{A}$.

Answer 1

Answer: (A), (B), (C), (D)

The magnitude of current flowing through $R_2$ is $1.2\text{A}$.

Given circuit as

Point $A\text{and}B$ are at same potential so they can be merged/folded.


For loop $1$

$-\frac{1}{2}x-\frac{1}{2}(x+y)-x+6=0$

$-2x-\frac{y}{2}=-6$

$-4x-y=-12$

$4x+y=12$ ...(1)

For loop $2$

$\frac{1}{2}y+1y+12+\frac{1}{2}(x+y)=0$

$\frac{3}{2}y+\frac{y}{2}+\frac{x}{2}=-12$

$2y+\frac{x}{2}=-12$

$4y+x=-24$ ...(2)
Now solving both the equations obtained above, we have

$4y+x-16x-4y$

$=-24-48$

$-15x=-72$

$x=\frac{72}{15}$

$4\left(\frac{72}{15}\right)+y=12$

$y=12-\frac{288}{15}$

$=\frac{180-288}{15}$

$=\frac{-108}{15}=-7.2\text{A}$

Current in $R_1=7.2\text{A}$

Current in $R_2=\frac{x+y}{2}=(\frac{72}{15}-\frac{108}{15})\frac{1}{2}$

$=\frac{1}{2}\left(\frac{36}{15}\right)=\frac{2.4}{2}\text{A}=1.2\text{A}$

Current in $R_3=x=4.8\text{A}$

Current in $R_5=\frac{1}{2}x=2.4\text{A}$.
Hence, all the options are correct.