Explaination :
Here the given circuit is both in series and parallel. So we would be first calculating the total resistance in series.
Rs = R1 + R2 .... n
Rs = 100Ω + 220Ω
Rs = 320Ω
Therefore, resistance in series combination is of 320Ω.
Now, the circuit would be combined in parallel.
As we know that,
- 1/Rp = 1/R1 + 1/R2 +... 1/n
>> 1/Rp = 1/320 + 1/320
>> 1/Rp = 1 + 1 / 320
>> 1/Rp = 2 / 320
>> 1/Rp = 1 / 160
Cross multiplying them,
>> Rp = 160Ω
Therefore,
- Equivalent resistance of circuit is of 160Ω
