Calculate the equivalent resistance of the network between terminals A and B.

anisingh259

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6Ω & 3Ω in parallel so the resultant of it will be -- 1/R= 1/6+1/3...... so its equivalent is 2Ω
2Ω & 6Ω is in series combination so the resultant R= 2 + 6 = 8Ω
8Ω & 8Ω is in parallel combination so the resultant 1/R= 1/8 + 1/8 ..... so its equivalent is 4Ω
4Ω & 4Ω(the resultant of 8Ω & 8Ω) is in series combination so the resultant R= 4 + 4 = 8Ω

now, (2Ω & 6Ω is in series combination so the resultant R= 2 + 6 = 8Ω) and (4Ω & 4Ω(the resultant of 8Ω & 8Ω) is in series combination so the resultant R= 4 + 4 = 8Ω) are in parallel combination .... so the resultant will be 4Ω

In last, (6Ω & 3Ω in parallel so the resultant of it will be -- 1/R= 1/6+1/3...... so its equivalent is 2Ω) and the resultant 4Ω in the last line will be in series combination and their resultant will be equal to 4Ω + 2Ω = 6Ω
Solution: 6Ω
 
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