Sol. (i) In the given circuit, series or parallel combinations are not formed. Therefore, to calculate equivalent resistance between terminals A and D, a battery of V volt is to be connected. If the circuit draws a current I from battery, its equivalent resistance will be equal to V/I.

When a battery is connected across terminals A and D, circuit becomes symmetric about a plane passing through mid points of edges BC and AD and vertex O. Therefore, current through different resistors will be as shown in Fig. .

Fig. Fig.

Since, current is AO and OD is equal, therefore, they may be assumed in series with each other. Similarly, BO and OC may be assumed in series. Hence, the circuit may be assumed as shown in Fig. . Therefore, equivalent resistance of each series combination (BO and OC and AO is in parallel with BC.)

Equivalent resistance of this combination = =

Now this combination is in series with resistors AB and CD. Therefore, equivalent resistance of this series combination = + 2r = r.

This series combination, resistors AD and series combination of resistors AO and OD are in parallel with each other. Hence, equivalent resistance R of the circuit is given by

= + + or R =

(ii) Now the battery is to be connected across terminals A and O and circuit becomes symmetric about a plane passing through edges AO and OC. Current through different resistors will be as shown in Fig. .

Applying Kirchhoff's voltage law on mesh OCDO.

I 2 r + r – I 1 r = 0.

or I 1 = I 2 ….. (1)

For mesh ODAO.

I 1 r + r – (I – 2I 1 – I 2 ) r = 0.

4I 1 + I 2 = I ….. (2)

From equations (1) and (2) I 1 = I and I 2 = I

For mesh OBA – V – O,

I 1 r + r – V = 0

Substituting values of I 1 and I 2 ,

Ir = V ….. (3)

But equivalent resistance of the circuit=

∴ Equivalent resistance =