# Which of the circuits illustrated below will give a total resistance of 1Ω…

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C

## DETAILS…

There’s no quicker way to determine the answer than to solve for the effective resistance of each option. Let’s do this:

### FOR OPTION A:

each of the resistors are in series. Effective resistance of resistors in series = addition of each resistance. Therefore,

effective resistance = R = R1 + R2 + R3 = 2 + 3 + 6 = 11Ω

### FOR OPTION B:

each of the resistors are in parallel. Effective resistance in parallel here will calculated thus:

1/R = 1/R1 + 1/R2

1/R = 1/2 + 1/3

1/R = (3 + 2)/6

1/R = 5/6. Hence, effective resistance = R = 6/5 = 1.2Ω

### FOR OPTION C:

the three resistors are in parallel. Effective resistance will then be calculated thus:

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/2 + 1/3 + 1/6

1/R = 1/2 + (1/3 + 1/6)

1/R = 1/2 + (2/6 + 1/6)

1/R = 1/2 + (3/6)

1/R = 1/2 + 1/2

1/R = 2/2 = 1

Hence, R = 1Ω

We have the answer. no need to resolve option D.

### Now for the right answer to the above question:

1. A is not the right answer. effective resistance here is 11Ω
2. B is incorrect. effective resistance here is 1.2Ω
3. C is correct. effective resistance here is 1Ω
4. D is incorrect. effective resistance here should be 4Ω

## KEY-POINTS…

• The quickest way to solve this would be to analyze each circuit beginning from the most suspected.
• Effective resistance of resistors in series is simply equal to addition of each.
• Effective resistance of parallel resistors is equal to the reciprocal of the added reciprocal of each.

You can also click the golden ASK A QUESTION BOTTON down this page to raise new related questions… OUR COMMUNITY OF EXPERTS WILL BE MORE THAN HAPPY TO TACKLE IT…

/ culled from 2020 JAMB-UTME physics past question 20 /