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#battery internal resistance

7 public questions tagged with this topic.

Electrical Power, Energy and Heating Effect

A \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance is connected to a \( 10 \, \Omega \) resistor.

**Heating effect** depends on I² R t, explaining why high currents cause significant heating, need for thick wires, fuses. Energy supplied by battery ε I t = I²(R+r)t, split between external and internal as per resistances. Total resistance: Rtₒtₐl = 10 + 2 = 12 Ω . Current: I = (ε/Rtₒtₐl) = (12/12) = 1 A . Power: P = I² r = 1² × 2 = 2 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 yields 2.0 W,

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

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Electrical Power, Energy and Heating Effect

A \( 6 \, \text{V} \) battery with \( 0.5 \, \Omega \) internal resistance is connected to a \( 5.5 \, \Omega \) resisto

**Heating effect** depends on I² R t, explaining why high currents cause significant heating, need for thick wires, fuses. Energy supplied by battery ε I t = I²(R+r)t, split between external and internal as per resistances. Total resistance: Rtₒtₐl = 5.5 + 0.5 = 6 Ω . Current: I = (ε/Rtₒtₐl) = (6/6) = 1 A . Power: P = I² R = 1² × 5.5 = 5.5 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 = I²R, evaluation yields 5.5 W,

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

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Resistance, Resistivity and Ohm's Law

A battery of emf \( 15 \, \text{V} \) and internal resistance \( 3 \, \Omega \) is connected to a resistor. If the termi

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. Voltage drop: I r = ε - V = 15 - 12 = 3 V . Current: I = (3/r) = (3/3) = 1 A . Resistance: R = (V/I) = (12/1) = 12 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 12 Ω,

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

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EMF, Internal Resistance and Cells Combination

Two cells of emf \( 2.5 \, \text{V} \) and \( 3.5 \, \text{V} \) with internal resistances \( 0.5 \, \Omega \) and \( 1

**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. For series: εₑq = ε₁ + ε₂ = 2.5 + 3.5 = 6.0 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 6.0 V,

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

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EMF, Internal Resistance and Cells Combination

A circuit has a \( 16 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and three resistors \( 2 \, \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. Parallel resistance: (1/R_p) = (1/2) + (1/4) + (1/8) = (4 + 2 + 1/8) = (7/8) ⇒ R_p = (8/7) ≈ 1.14 Ω . Total resistance: Rtₒtₐl = 2 + 1.14 = 3.14 Ω . Total current: I = (ε/Rtₒtₐl) = (16/3.14) ≈ 5.10 A . Voltage across

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

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EMF, Internal Resistance and Cells Combination

A cell of emf \( 7 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 6 \, \Omega \) 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 = 6 + 1 = 7 Ω . Current: I = (ε/Rtₒtₐl) = (7/7) = 1 A . Terminal voltage: V = ε - I r = 7 - 1 × 1 = 6 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

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

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EMF, Internal Resistance and Cells Combination

A circuit has a \( 28 \, \text{V} \) battery with \( 4 \, \Omega \) internal resistance and three resistors \( 4 \, \Ome

**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. Parallel resistance: (1/R_p) = (1/4) + (1/8) + (1/16) = (4 + 2 + 1/16) = (7/16) ⇒ R_p = (16/7) ≈ 2.29 Ω . Total resistance: Rtₒtₐl = 4 + 2.29 = 6.29 Ω . Total current: I = (ε/Rtₒtₐl) = (28/6.29) ≈ 4.45 A . Voltage across parallel: V = I R_p = 4.45 × 2.29 ≈ 10.19 V . Current through

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

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