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Question
Determine the equivalent resistance of networks shown in Fig.
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Solution
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.
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