Question

For the circuit shown in figure, the equivalent resistance between $\text{A}$ and $\text{C}$ will be

• A. $\frac{12r}{11}$
• B. $\frac{13r}{11}$
• C. $\frac{14r}{11}$​​​​
• D. $\frac{15r}{11}$​​

Answer 1

The correct option is D $\frac{15r}{11}$​​

Let us connect an imaginary battery of $V\text{Volt}$ between points $\text{A}$ and $\text{C}$.

Due to symmetric arrangements of resistances with respect to point $\text{A}$ and $\text{C}$, the current distribution will be as shown in figure.


Applying $KCL$ at junction $\text{E}$ or $\text{F}$ gives,

$i_2-i_3=i_1+i_3+i_4$

$\Rightarrow i_1-i_2+2i_3+i_4=0........\left(1\right)$

Applying $KVL$ in closed loop $\text{BFAB}$ or $\text{ECDE}$, we get,

$-r{i}_3+i_{1}r-i_{2}r=0$

$\Rightarrow i_1=i_2+i_3........\left(2\right)$

For path $\text{ADC}$, we can write potential difference as,

$V_A-i_{1}r-(i_2-i_3)r-i_{2}r=V_C$

$V_A-V_C=i_{1}r+i_{2}r+(i_2-i_3)r$

$\Rightarrow V=i_{1}r+2i_{2}r-i_{3}r.....\left(3\right)$

From Eq.(2) & (3),

$V=(i_2+i_3)r+2i_{2}r-i_{3}r$

$V=3i_{2}r$

$\therefore i_2=\frac{V}{3r}$

Similarly, we can get $i_1=\frac{2V}{5r}$

Now total current supplied by battery.

$i=i_1+i_2=\frac{2V}{5r}+\frac{V}{3r}$

$\Rightarrow i=\frac{11V}{15r}$

Let $R_{eq}$ be net resistance of circuit across $\text{A}$ and $\text{C}$. So, current will be

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

$\Rightarrow R_{eq}=\frac{V}{i}=\frac{V}{\left(\frac{11V}{15r}\right)}$

$\therefore R_{eq}=\frac{15r}{11}$

Therefore, option $\left(d\right)$ is correct.

Why this question? Tip: In problems where line of symmetry is not found, but resistance are arranged in some symmetrical manner w.r.t end point / terminals. Assume a battery connected & apply $KVL$ & $KCL$ by showing current distribution.