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Question

Why does the equivalent resistance of two resistors in parallel always lie between zero and the
smallest individual resistance?

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Explanation

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. For parallel resistors, 1/Rₑq = 1/R₁ + 1/R₂ , so Rₑq = R₁ R₂ / (R₁ + R₂) . This value is less than the smaller resistance (e.g., if R₁ < R₂ , Rₑq < R₁ ) but greater than zero, as additional paths reduce resistance. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and

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