In a parallel combination of resistors, the total resistance of the circuit is the individual resistances.

less than

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Solution

The correct option is A less than
When resistors are connected in parallel, the equivalent resistance is less than the lowest resistance in the circuit.
Proof is as follows:
As we know when $R_{1}, R_{2}, R_{3} . . . R_{n}$ resistors are connected in parallel circuit, the equivalent resistance of the circuit is
$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} . . . \frac{1}{R_{n}}$
$\frac{1}{R_{e q}} \geq \frac{1}{R_{i}} \forall \in {1, 2, 3, . . . ., n}$
$⟹ R_{e q} \leq R_{i}$

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Similar questions

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