Question

Determine the equivalent resistance of networks shown in Fig.

Answer 1

It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in series.

Hence, their equivalent resistance = (1 + 1) = 2 Ω

It can also be observed that two resistors of resistance 2 Ω each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

Therefore, the circuit can be redrawn as

It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all four loops. Hence, equivalent resistance (R’) of each loop is given by,

R' = `(2 xx 4)/(2 + 4)`

= `8/6`

= `4/3` Ω

The circuit reduces to,

All four resistors are connected in series.

Hence, the equivalent resistance of the given circuit is `4/3 xx 4` = `16/3` Ω.