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#circuit problem

3 public questions tagged with this topic.

Potentiometer, Conductivity and Special Cases

A circuit has a \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 5 \, \Omega

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. Parallel resistance: (1/R_p) = (1/5) + (1/10) = (2 + 1/10) = (3/10) ⇒ R_p = (10/3) ≈ 3.33 Ω . Total resistance: Rtₒtₐl = 1 + 3.33 = 4.33 Ω . Current: I = (ε/Rtₒtₐl) = (10/4.33) ≈ 2.31 A . 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

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Electric Current, Drift Velocity and Mobility

A \( 15 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Terminal voltage: V = ε - I r = 15 - 2 × 3 = 9 V . Resistance: R = (V/I) = (9/2) = 4.5 Ω . 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 4.5

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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EMF, Internal Resistance and Cells Combination

A \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 9 \, \Omega \) resistor.

**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Total resistance: Rtₒtₐl = 9 + 1 = 10 Ω . Current: I = (ε/Rtₒtₐl) = (10/10) = 1 A . Power: P = I² r = 1² × 1 = 1 W . 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

Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination

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