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#terminal voltage

8 public questions tagged with this topic.

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

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

Why does the terminal voltage of a battery become equal to its emf when no current flows?

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Terminal voltage V = ε - I r . When I = 0 (open circuit), the internal voltage drop I r = 0 , so V = ε , matching the emf. 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 No internal voltage drop,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

A cell of emf \( 8 \, \text{V} \) and internal resistance \( 2 \, \Omega \) is connected to a \( 6 \, \Omega \) resistor

**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. Total resistance: Rtₒtₐl = 6 + 2 = 8 Ω . Current: I = (ε/Rtₒtₐl) = (8/8) = 1 A . Terminal voltage: V = ε - I r = 8 - 1 × 2 = 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 and P =

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

A cell of emf \( 6 \, \text{V} \) and internal resistance \( 2 \, \Omega \) is connected to an external resistor \( 4 \,

**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. Current: I = (ε/R + r) = (6/4 + 2) = 1 A . Terminal voltage: V = ε - I r = 6 - 1 × 2 = 4 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 4 V,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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

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

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

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