EMF, Internal Resistance and Cells Combination

EMF, Internal Resistance and Cells Combination

A cell of emf \( 6 \, \text{V} \) and internal resistance \( 1.5 \, \Omega \) is connected to a \( 4.5 \, \Omega \) resi

**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: Rtₒtₐl = 4.5 + 1.5 = 6 Ω . Current: I = (ε/Rtₒtₐl) = (6/6) = 1 A . Terminal voltage: V = ε - I r = 6 - 1 × 1.5 = 4.5 V . Applying I = n e A v_d, R = ρ

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

View Details

EMF, Internal Resistance and Cells Combination

A cell of emf \( 10 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 9 \, \Omega \) resisto

**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 . Terminal voltage: V = ε - I r = 10 - 1 × 1 = 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

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

View Details

EMF, Internal Resistance and Cells Combination

A \( 18 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 18 - 2 × 3 = 12 V . Resistance: R = (V/I) = (12/2) = 6 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

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

View Details

EMF, Internal Resistance and Cells Combination

A \( 10 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**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. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 1.5 = 1.25 Ω . Total current: Itₒtₐl = (V/Rₑq) = (10/1.25) = 8 A . Corner current: I = (Itₒtₐl/3) = (8/3) ≈ 2.67 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 - I r and P =

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

View Details

EMF, Internal Resistance and Cells Combination

Two cells in parallel have emf \( 12 \, \text{V} \) and \( 3 \, \text{V} \) with internal resistances \( 4 \, \Omega \)

**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 parallel: εₑq = (ε₁ r₂ + ε₂ r₁/r₁ + r₂) . Substitute: εₑq = (12 × 1 + 3 × 4/4 + 1) = (12 + 12/5) = (24/5) = 4.8 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

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

View Details

EMF, Internal Resistance and Cells Combination

Why does a battery’s emf remain constant even when it delivers a large current?

**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. Emf is the potential difference across the battery terminals when no current flows, determined by the chemical reactions inside. It’s an intrinsic property, independent of current, though terminal voltage drops due to internal resistance. 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

View Details

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

View Details

EMF, Internal Resistance and Cells Combination

Two cells of emf \( 4 \, \text{V} \) and \( 5 \, \text{V} \) with internal resistances \( 0.5 \, \Omega \) and \( 1 \, \

**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. Equivalent emf: εₑq = 4 + 5 = 9 V . Total resistance: Rtₒtₐl = 0.5 + 1 + 5.5 = 7 Ω . Current: I = (εₑq/Rtₒtₐl) = (9/7) ≈ 1.29 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 - I r and P = I²R, evaluation

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

View Details

EMF, Internal Resistance and Cells Combination

Why does a conductor’s current density increase when its length is halved while keeping the potential difference constan

**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. Resistance R = rho l / A . Halving length ( l' = l/2 ) halves R ( R' = R/2 ). Current I = V / R , so I' = V / (R/2) = 2I . Current density j = I / A , so j' = 2I / A = 2j . Applying I = n

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

View Details

EMF, Internal Resistance and Cells Combination

A \( 14 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 14 - 2 × 2 = 10 V . Resistance: R = (V/I) = (10/2) = 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 =

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

View Details

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

View Details

EMF, Internal Resistance and Cells Combination

A \( 8 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 7 \, \Omega \) resistor. W

**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: Rtₒtₐl = 7 + 1 = 8 Ω . Current: I = (ε/Rtₒtₐl) = (8/8) = 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

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

View Details