Question

The equivalent resistance between the points A and B of the circuit shown in figure is.

• A. $7\Omega$
• B. $5\Omega$
• C. $\frac{5}{7}\Psi$
• D. $\frac{7}{5}\Psi$

Answer 1

The correct option is A $7\Omega$

This is clearly not a Wheatstone bridge as $\frac{R_1}{R_2}\ne \frac{R_3}{R_4}$

Which in turn implies that points C and D are not at the same potential and there will be a current through the 5$\Omega$ resistor.
Let’s simplify this and connect a battery of V volts across A and B. Also let’s say a total current of i is provided by the battery. Refer the given diagram below:

SYMMETRY ARGUMENTS:
If I modify the circuit, so that, the point B is at the place of A now and vice-versa. Would you be able to differentiate? No right!
This is called point symmetry. It also implies that the current distribution should also be symmetric, which simply means this-

If you are still not clear, you can see from the above diagram that the distribution of current remains the same while leaving the branches in the same order as they entered. This means that there will be a current of $i_1-i_2$ through CD as shown below:

Now the question is simplified. We can use Kirchhoff’s law to calculate $i_1$ and $i_2$ in terms of V and apply $V=i{R}_{eq}$ to get $R_{eq}$.
Let’s do that. Refer the following diagram:

Loop ACDFA
$-10i_1-5(i_1-i_2)+5i_2=0-15i_1+10i_2=0i_1=\frac{2}{3}{i}_2$
Loop ACEBGHA:
$-10i_1-5i_2+V=010i_1+5i_2=V$
Solving both the equations in finding $i_1$ and $i_2$ in terms of V, we get
$i_2=\frac{3}{35}V$
$i_1=\frac{2}{35}V$
$i=i_1+i_2=\frac{5}{35}V=\frac{1}{7}V$
Which implies, $\frac{V}{i}=7\Omega$
Thus equivalent resistance between A and B is $7\Omega$.