#load resistor

Rectifiers, Filters and Applications

In a rectifier with a capacitor filter, the capacitor discharges through:

**Rectifier applications** diode must have reverse breakdown voltage higher than peak inverse voltage, centre-tap transformer provides two opposite phase voltages for full-wave, capacitor discharges through load R_L when diode off, drift current in junction is minority carrier motion due to field, diffusion due to gradient, diode conducts when forward biased anode positive. The capacitor charges to the peak voltage and discharges through the load resistor ( R_L ) during the non-conducting half-cycle, smoothing the output voltage. Substituting values gives Load resistor, which matches expected

Ref: NCERT > Physics Book > Electronic Devices > Rectifiers, Filters and Applications

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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,

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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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

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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

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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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