Three resistors with magnitudes 2, 4 and 8 ohm are connected in parall

Three resistors with magnitudes 2, 4 and 8 ohm are connected in parallel. The equivalent resistance of the system would be

less than 2 ohm
more than 2 ohm but less than 4 ohm
4 ohm
14 ohm
This question was previously asked in
UPSC NDA-1 – 2016
The correct option is A.
When resistors are connected in parallel, the reciprocal of the equivalent resistance (Req) is equal to the sum of the reciprocals of the individual resistances.
Given resistances are R₁ = 2 ohm, R₂ = 4 ohm, and R₃ = 8 ohm.
The formula for equivalent resistance in parallel is:
1/Req = 1/R₁ + 1/R₂ + 1/R₃
Substituting the values:
1/Req = 1/2 + 1/4 + 1/8
To add these fractions, find a common denominator, which is 8:
1/Req = 4/8 + 2/8 + 1/8
1/Req = (4 + 2 + 1) / 8
1/Req = 7/8
Now, take the reciprocal to find Req:
Req = 8/7 ohm
To compare this value with the options, calculate the decimal value: Req ≈ 1.14 ohm.
A key principle of parallel resistance is that the equivalent resistance is *always less than the smallest individual resistance* in the combination. In this case, the smallest resistance is 2 ohm, and 8/7 ohm (approx 1.14 ohm) is indeed less than 2 ohm.
Connecting resistors in parallel provides multiple paths for the current to flow, effectively reducing the overall resistance of the circuit. Connecting resistors in series adds their resistances, resulting in a total resistance greater than any individual resistance.