Resistor network with mirror symmetry

JEE Advanced 2012 Paper 1, Question 12

For the resistance network shown in the figure, choose the correct option(s).

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  1. The current through PQ is zero.
  2. I_1 = 3 A.
  3. The potential at S is less than that at Q.
  4. I_2 = 2 A.


We can speed up the solution by noticing the mirror symmetry of the circuit. Let the current I_1 split into two parts I_a and I_b at the node X (see figure). Similarly, let the currents going into node Y be I_a' and I_b'. If we were to reverse the polarity of the 12 \, {\rm V} battery connected to the circuit, all currents would simply reverse direction, but their magnitudes would remain the same. But then, the currents leaving the node Y must be I_a and I_b since, from the battery’s perspective, the circuit looks the same under X \leftrightarrow Y. Thus, I_a'=I_a and I_b'=I_b.

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By similar reasoning the currents through the 1 \Omega resistors must be the same, and we label these I_\Delta. Using Kirchoff’s current law we can fill in the remaining currents through PS and QT (see figure below).

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Applying Kirchoff’s voltage law on the loop XPQ,

Similarly, the loop PQTS gives

(1)   \begin{equation*}   -2 (I_a + I_\Delta) - I_\Delta + 4 (I_b - I_\Delta) - I_\Delta = 0 \end{equation*}

Clearly, both of these equations can only be satisfied if I_\Delta = 0. So option (A) is correct.

Since no current flows through the 1 \Omega resistors and we can treat the legs XPSY and XQTY as independent. Then,

(2)   \begin{equation*}   I_2 = I_a = \frac{12 \, \rm{V}}{6 \, \Omega} = 2 \, {\rm A}   \implies   \text{ \bf option (D) is correct} , \end{equation*}

and the total current I_1 is

(3)   \begin{equation*}   I_1 = I_a + I_b = \frac{12 \, \rm{V}}{6 \, \Omega} + \frac{12 \, \rm{V}}{12 \, \Omega} = 3 \, {\rm A}   \implies   \text{ \bf option (B) is correct} , \end{equation*}

The potential at point S is V_S = 2 I_a = 4 \, {\rm V}, and that at point Q is V_Q = 8 I_b = 8 \, {\rm V}, and so V_S < V_Q which means option (C) is correct.

This, all the given options are correct.

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