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Symmetry can be used to solve certain electrical circuits.

In electrical circuits, if the components and their arrangement are identical on either side of a central axis, the circuit is said to have mirror symmetry. Symmetrical circuits have the property that the voltage and current relationships at corresponding points are equal, making analysis and problem-solving easier.

**Problem:** Find the equivalent resistance between cube's main diagonal.

**Problem:** Find the equivalent resistance between A and B. Answer is 4R/5.

**Problem:** Find the equivalent resistance between A and B. Answer is 3R/2.

**Problem:** Find the equivalent resistance between A and B. Answer is 6R/7.

**Problem:** Find the equivalent resistance between (i) A and D and (ii) B and C. Answer is $R_\text{AD}=R$.

**Solution:**
From symmetry, we see that all nodes at a given level n are equipotential points. We can add dummy wires between these equipotential points without changing the current flows in the circuit. The equivalent circuit has n levels in series, with $2^{n−1}$ resistors in parallel at the n-th level.

The resistance of the n-th level is ($2^{n-1}$ resistors in parallel) is \begin{align} R_n=\frac{R}{2^{n-1}}. \end{align} The resistance of a pyramid of N levels is the sum of the resistances $R_n$ from $n = 1$ to $N$ i.e., \begin{align} R_N&=\sum_{n=1}^{N}\frac{R}{2^{n-1}}\\ &=R\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{N-1}}\right). \end{align} The sum of this geometric series is \begin{align} R_N=2R\left[1-\left(\frac{1}{2}\right)^N\right]. \end{align} Take limit $N\to\infty$ to get resistance of the infinite resistance \begin{align} R_\infty=\lim_{N\to\infty}2R\left[1-\left(\frac{1}{2}\right)^N\right]=2R. \end{align}

**Aliter:**
We can also think about this problem in terms of current symmetry at each level. At the n-th level, there are $2^{n−1}$ resistors, each with the same current (due to the symmetry). If the current through the entire pyramid is I, then the current through each resistor in the n-th level is $I_n = I/2^{n−1}$. We need to add the voltage drop across eachlevel in series to find the voltage drop across the entire pyramid. The potential drop across a resistor in the n-th level is
\begin{align}
V_n = I_nR = IR/2^{n−1}.
\end{align}
So the total voltage drop across a pyramid of N levels is
\begin{align}
V_N & =\sum_{n=1}^{N}\frac{IR}{2^{n-1}} \\
&=IR\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdot+\frac{1}{2^{N-1}}\right)\\
&=2IR\left[1-\left(\frac{1}{2}\right)^N\right].
\end{align}
The resistance of a pyramid of N levels is given by $R_N = V_N /I$ i.e.,
\begin{align}
R_N=2R\left[1-\left(\frac{1}{2}\right)^N\right].
\end{align}
Take limit $N\to\infty$ to get resistance of the infinite resistance
\begin{align}
R_\infty=\lim_{N\to\infty}2R\left[1-\left(\frac{1}{2}\right)^N\right]=2R.
\end{align}

**Problem (IIT JEE 2002): **
The effective resistance between points P and Q of the electrical circuit shown in the figure is__________

**Solution: **
The given circuit is a good example of symmetry. When looking from P, branch PAB looks same as the branch PFE and hence the same current $i_1$ should flow through these branches.

Similarly, when looking from O, branches OB and OE look identical and same current $i_3$ should flow through them. Branches through P and Q are also identical. (Similar symmetry arguments can be used at other points in the circuits.) Apply Kirchhoff's law in loop ABOPA and BCQOB to get \begin{alignat}{2} &2Ri_1+2Ri_3-i_2r=0, \\ &2R(i_1-i_3)-r(i_2+2i_3)-2Ri_3=0 . \end{alignat} Solve above equations to get $i_3=0$ i.e., no current flows in the branch OB and OE. Hence, branch OB and OE can be removed from the circuit without affecting the effective resistance.

Thus, effective resistance of the circuit is \begin{align} R_e=(4R\parallel4R)\parallel 2r=\frac{2Rr}{R+r}. \end{align} We encourage you to exploit the symmetry to check that $i_3=0$.

**Problem (JEE Mains 2017): **
In the given circuit, the current in each resistance is

- 1 A
- 0.25 A
- 0.5 A
- 0 A

**Hint:**
use symmetry (or Kirchhoff's law) to argue that current is zero in each branch.

**Problem (JEE 2012): **
For the resistance network shown in the figure, choose the correct option(s),

- The current through PQ is zero.
- $i_1=3$ A
- The potential at S is less than that at Q.
- $i_2=2$ A