Question
Determine the equivalent resistance of networks shown in Fig.
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Answer
- It can be observed from the given circuit that in the first small loop, two resistors of resistance $1\ \Omega$ each are connected in series.
It can also be observed that two resistors of resistance $2\ \Omega$ each are connected in series.
Hence, their equivalent resistance $=(2+2)=4\ \Omega$
Therefore, the circuit can be redrawn as
It can be observed that $2\ \Omega\ \text{and}\ 4\ \Omega$ resistors are connected in parallel in all the four loops. Hence, equivalent resistance (R') of each loop is given by,
$\text{R}=\frac{2\times4}{2+4}=\frac{8}{6}=\frac{4}{3}\Omega$
The circuit reduces to,
All the four resistors are connected in series.
Hence, equivalent resistance of the given circuit is $\frac{4}{3}\times4=\frac{16}{3}\Omega$
- It can be observed from the given circuit that five resistors of resistance R each are connected in series.
= 5 R
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