#resistor current

Potentiometer, Conductivity and Special Cases

Why does the power dissipated in a resistor increase quadratically with current?

**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Power P = I² R . Since power depends on the square of the current ( I² ), doubling the current quadruples the power, assuming resistance remains constant. 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 Power depends on current squared,

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

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Kirchhoff's Laws and Combination of Resistors

In a circuit with resistors in series, why does the current remain the same through each resistor?

**Total voltage drop** across series equals source voltage because loop rule Σ V = ε, with internal resistance r included V = ε - I r. For 4 Ω,8 Ω,16 Ω parallel, 1/R_p =1/4+1/8+1/16=7/16, R_p=16/7≈2.29 Ω, then total resistance with internal 3 Ω is 5.29 Ω, current I=18/5.29≈3.4 A. In series, there’s only one path for current. Kirchhoff’s junction rule ensures charge conservation, so the same current flows through each resistor as no charge accumulates. 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 > Kirchhoff's Laws and Combination of Resistors

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and three resistors \( 3 \, \Ome

**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. Parallel resistance: (1/R_p) = (1/3) + (1/6) + (1/9) = (6 + 3 + 2/18) = (11/18) ⇒ R_p = (18/11) ≈ 1.64 Ω . Total resistance: Rtₒtₐl = 2 + 1.64 = 3.64 Ω . Total current: I = (ε/Rtₒtₐl) = (12/3.64) ≈ 3.3 A . Voltage across parallel: V = I R_p = 3.3 × 1.64 ≈ 5.41 V . Current through 9 Ω :

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 30 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three resistors \( 3 \, \Ome

**Kirchhoff's loop rule** (energy conservation) states algebraic sum of potential differences around closed loop zero, Σ ε - Σ I R =0, ensures total voltage drop across series resistors equals source emf. Junction rule (charge conservation) Σ I_in = Σ I_out ensures current continuity. Parallel resistance: (1/R_p) = (1/3) + (1/6) + (1/12) = (4 + 2 + 1/12) = (7/12) ⇒ R_p = (12/7) ≈ 1.71 Ω . Total resistance: Rtₒtₐl = 3 + 1.71 = 4.71 Ω . Total current: I = (ε/Rtₒtₐl) = (30/4.71) ≈ 6.37 A . Voltage across parallel: V = I R_p = 6.37 × 1.71 ≈ 10.89 V

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

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