#resistors

AC Fundamentals - RMS, Average and Peak Values

A \( 198.1 \, \text{V} \) (peak) AC source is connected to a \( 70 \, \Omega \) resistor. What is the average power cons

**Peak current** I_peak = V_peak/R for resistor, I_rms = V_rms/R, V_peak = √2 V_rms, for 200 V rms, V_peak=282.8 V, I_peak=282.8/80=3.535 A, rms I=200/80=2.5 A, average over complete cycle zero because positive and negative halves cancel. RMS voltage: V = (v_m/√(2)) = (198.1/1.414) ≈ 140 V . RMS current: I = (V/R) = (140/70) = 2 A . Average power: P = I² R = 2² × 70 = 280 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 280 W, consistent with phasor analysis

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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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

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

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 10 \, \Omega \), \( R_2 = 20 \, \Omega \), \( R_3 = 15 \, \Omega \), \( R_4 = 30 \, \O

**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. Check balance: (R₁/R₂) = (10/20) = 0.5 , (R₃/R₄) = (15/30) = 0.5 . Bridge is balanced. Since balanced, I_g = 0 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 yields 0 A,

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

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EMF, Internal Resistance and Cells Combination

A \( 18 \, \text{V} \) 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) × 3 = 2.5 Ω . Total current: I = (V/Rₑq) = (18/2.5) = 7.2 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 series/parallel, V = ε - I r and P = I²R, evaluation yields 7.2 A,

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

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 16 \, \Omega \), \( R_2 = 32 \, \Omega \), \( R_3 = 24 \, \Omega \), \( R_4 = 48 \, \O

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Check balance: (R₁/R₂) = (16/32) = 0.5 , (R₃/R₄) = (24/48) = 0.5 . Bridge is balanced. Since balanced, I_g = 0 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 yields 0 A,

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

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 4 \, \Omega \), \( R_2 = 8 \, \Omega \), \( R_3 = 6 \, \Omega \), \( R_4 = 12 \, \Omeg

**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. Check balance: (R₁/R₂) = (4/8) = 0.5 , (R₃/R₄) = (6/12) = 0.5 . Bridge is balanced. Since balanced, I_g = 0 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 yields 0 A,

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

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