#series resistors

Potentiometer, Conductivity and Special Cases

A \( 12 \, \text{V} \) battery with negligible internal resistance is connected to a \( 4 \, \Omega \) and \( 8 \, \Omeg

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Total resistance: R = 4 + 8 = 12 Ω . Current: I = (V/R) = (12/12) = 1 A . Power: P = I² R = 1² × 4 = 4 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

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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Electrical Power, Energy and Heating Effect

A \( 15 \, \text{V} \) battery with negligible internal resistance is connected to a \( 5 \, \Omega \) and \( 10 \, \Ome

**Electrical power** P = V I = I² R = V²/R (W), energy E = P t = I² R t (J), heating effect Joule's law H = I² R t. When internal r equals external R, total resistance 2R, I = ε/2R, power in external = I²R = ε²/4R, total = ε²/2R, fraction external = 1/2, illustrating maximum power transfer when R = r. Total resistance: R = 5 + 10 = 15 Ω . Current: I = (V/R) = (15/15) = 1 A . Power: P = I² R = 1² × 5 = 5 W . Applying I = n e A

Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect

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Resistance, Resistivity and Ohm's Law

In a circuit with two resistors in series, why does the voltage divide proportionally to their resistances?

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Voltage drop across each resistor is V = I R . With the same current in series, V ∝ R , so the total voltage splits in proportion to the resistances (e.g., V₁ / V₂ = R₁ / R₂ ). Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0,

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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EMF, Internal Resistance and Cells Combination

A \( 15 \, \text{V} \) battery with negligible internal resistance is connected to a \( 3 \, \Omega \) and \( 9 \, \Omeg

**EMF ε** is work done by non-electrostatic forces per unit charge, terminal voltage V = ε - I r, r internal resistance (Ω), I current (A). When external R = r, total resistance 2r, current I = ε/2r, power in external R is I²R = ε²/4r, total power ε²/2r, so half power dissipated externally, half internally. Total resistance: R = 3 + 9 = 12 Ω . Current: I = (V/R) = (15/12) = 1.25 A . Power: P = I² R = (1.25)² × 9 = 1.5625 × 9 = 14.06 W ≈ 14 W . Applying I = n e A v_d, R

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

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EMF, Internal Resistance and Cells Combination

A \( 20 \, \text{V} \) battery with negligible internal resistance is connected to a \( 5 \, \Omega \) and \( 15 \, \Ome

**EMF ε** is work done by non-electrostatic forces per unit charge, terminal voltage V = ε - I r, r internal resistance (Ω), I current (A). When external R = r, total resistance 2r, current I = ε/2r, power in external R is I²R = ε²/4r, total power ε²/2r, so half power dissipated externally, half internally. Total resistance: R = 5 + 15 = 20 Ω . Current: I = (V/R) = (20/20) = 1 A . Power: P = I² R = 1² × 15 = 15 W . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's

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

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EMF, Internal Resistance and Cells Combination

A \( 18 \, \text{V} \) battery with negligible internal resistance is connected to a \( 6 \, \Omega \) and \( 12 \, \Ome

**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Total resistance: R = 6 + 12 = 18 Ω . Current: I = (V/R) = (18/18) = 1 A . Power: P = I² R = 1² × 6 = 6 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

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

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