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#internal resistance

53 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

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Potentiometer, Conductivity and Special Cases

In a circuit with a battery, why does the internal resistance of the battery affect the maximum power delivered to an ex

**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. Maximum power transfer occurs when the external resistance equals the internal resistance ( R = r ), as P = I² R = (ε / (R + r))² R . Internal resistance limits current, influencing the power distribution. 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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Potentiometer, Conductivity and Special Cases

What is the primary reason a battery’s terminal voltage is less than its emf when supplying current?

**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. Terminal voltage V = ε - I r . When current flows, the voltage drop across the internal resistance ( I r ) reduces the voltage available at the terminals below the emf ( ε ). 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 =

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Potentiometer, Conductivity and Special Cases

A cell of emf \( 9 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 8 \, \Omega \) resistor

**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. Total resistance: Rtₒtₐl = 8 + 1 = 9 Ω . Current: I = (ε/Rtₒtₐl) = (9/9) = 1 A . Terminal voltage: V = ε - I r = 9 - 1 × 1 = 8 V . 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

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Wheatstone Bridge and Meter Bridge

A \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 3 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 12 - 3 × 2 = 6 V . Resistance: R = (V/I) = (6/3) = 2 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 2.0 Ω,

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Temperature Dependence of Resistance and Resistivity

Two cells in parallel have emf \( 15 \, \text{V} \) and \( 5 \, \text{V} \) with internal resistances \( 3 \, \Omega \)

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. For parallel: (1/rₑq) = (1/r₁) + (1/r₂) = (1/3) + (1/1) = (1 + 3/3) = (4/3) . rₑq = (3/4) = 0.75 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 0.75 Ω,

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Temperature Dependence of Resistance and Resistivity

Two cells in parallel have emf \( 10 \, \text{V} \) and \( 4 \, \text{V} \) with internal resistances \( 5 \, \Omega \)

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. For parallel: εₑq = (ε₁ r₂ + ε₂ r₁/r₁ + r₂) . Substitute: εₑq = (10 × 2 + 4 × 5/5 + 2) = (20 + 20/7) = (40/7) ≈ 5.71 V . 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 5.71 V,

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Temperature Dependence of Resistance and Resistivity

Two cells in parallel have emf \( 8 \, \text{V} \) and \( 4 \, \text{V} \) with internal resistances \( 2 \, \Omega \) a

**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. For parallel: (1/rₑq) = (1/r₁) + (1/r₂) = (1/2) + (1/1) = (1 + 2/2) = (3/2) . rₑq = (2/3) ≈ 0.67 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 0.67 Ω,

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Temperature Dependence of Resistance and Resistivity

Two cells of emf \( 3 \, \text{V} \) and \( 6 \, \text{V} \) with internal resistances \( 1 \, \Omega \) and \( 2 \, \Om

**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. For series: εₑq = ε₁ + ε₂ = 3 + 6 = 9 V . 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 9.0 V,

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Electrical Power, Energy and Heating Effect

A \( 9 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 8 \, \Omega \) resistor. W

**Power dissipation** in resistor converts electrical energy to heat, P = V²/R inversely proportional to R for fixed V, directly proportional for fixed I. For battery with internal r, power wasted internally = I² r, useful power = I² R, efficiency η = R/(R+r). Total resistance: Rtₒtₐl = 8 + 1 = 9 Ω . Current: I = (ε/Rtₒtₐl) = (9/9) = 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

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Electrical Power, Energy and Heating Effect

A \( 9 \, \text{V} \) battery with \( 0.5 \, \Omega \) internal resistance is connected to a \( 8.5 \, \Omega \) resisto

**Power dissipation** in resistor converts electrical energy to heat, P = V²/R inversely proportional to R for fixed V, directly proportional for fixed I. For battery with internal r, power wasted internally = I² r, useful power = I² R, efficiency η = R/(R+r). Total resistance: Rtₒtₐl = 8.5 + 0.5 = 9 Ω . Current: I = (ε/Rtₒtₐl) = (9/9) = 1 A . Power: P = I² R = 1² × 8.5 = 8.5 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

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

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 8 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two resistors \( 4 \, \Omega

**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. Parallel resistance: (1/R_p) = (1/4) + (1/4) = (2/4) = 0.5 ⇒ R_p = 2 Ω . Total resistance: Rtₒtₐl = 2 + 2 = 4 Ω . Current: I = (ε/Rtₒtₐl) = (8/4) = 2 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

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