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Question

Why does the resistivity of an insulator decrease dramatically when impurities are added?

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Explanation

**Wheatstone bridge balance** condition R₁/R₂ = R₃/R₄, R₄ = R₂ R₃/R₁, when galvanometer current zero, potentials at midpoints equal. At balance, no current through galvanometer, enabling precise resistance measurement independent of source voltage. Impurities in insulators (e.g., semiconductors) introduce additional charge carriers by creating energy levels within the bandgap, allowing more electrons or holes to conduct, reducing resistivity. 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 P = I²R, evaluation yields Charge carrier density increases,

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