Question

The equivalent resistance between points A and B of the frame formed by nine identical wires each of resistance R as shown in figure is

• A. $\frac{2}{11}R$
• B. $\frac{9}{11}R$
• C. $\frac{15}{11}R$
• D. $\frac{1}{11}R$

Answer 1

Answer: (C)

$\frac{15}{11}R$

Consider the circuit as shown in figure.

According to Kirchhoff's current law:

$I_1=I_3+I_5$
$I_2+I_3=I_4+I_5$

Also, the potential between AB is:
$(I_2+I_5+I_1)R=V$
And $(I_3+I_4)R=I_{5}R$
$(I_1+I_3)R=I_{2}R$

Therefore,
$I_2=\frac{6}{5}{I}_1$
$I_3=\frac{I_1}{5}$
$I_4=\frac{3}{5}{I}_1$
$I_5=\frac{4}{5}{I}_1$

$\therefore V=3I_{1}R$

But, $R_{AB}=\frac{V}{I_1+I_2}$
Hence, $R_{AB}=\frac{V}{I_1+\left(\frac{6}{5}\right)I_1}$
$R_{AB}=\frac{5V}{11I_1}$

Using value of $V$ we get,
$R_{AB}=\frac{5\left(3I_{1}R\right)}{11I_1}$
$R_{AB}=\frac{15R}{11}$