#electrical calculations

AC Through Resistor - Phasor and Power

A \( 190 \, \text{V} \) (rms) source supplies a \( 95 \, \Omega \) resistor. What is the peak current?

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. RMS current: I = (V/R) = (190/95) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 A . 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 2.828 A, consistent

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

View Details

AC Through Resistor - Phasor and Power

A \( 200 \, \text{V} \) (rms) AC source supplies a \( 100 \, \Omega \) resistor. What is the peak voltage?

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. Peak voltage: v_m = √(2) V . V = 200 V . v_m = 1.414 × 200 = 282.8 V ≈ 283 V . 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 283 V, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

View Details

AC Through Capacitor - Capacitive Reactance

A \( 60 \, \Omega \) resistor is connected to a \( 180 \, \text{V} \) (rms) AC source. What is the rms current?

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. RMS current: I = (V/R) . Given: V = 180 V , R = 60 Ω . I = (180/60) = 3 A . 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 3 A, consistent with phasor analysis

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

View Details

AC Through Inductor - Inductive Reactance

A \( 15 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the r

**AC through inductor** voltage leads current by 90°, V = L dI/dt, V(t)=V_peak sin(ωt+90°), I(t)=I_peak sin ωt, instantaneous power P= V I =½ V_peak I_peak sin2ωt, average zero over cycle because energy stored in magnetic field ½ L I² returned to source each quarter cycle, no net dissipation. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 15 × 10⁻⁶ F . X_C = (1/376.8 × 15 × 10⁻⁶) ≈ 177 Ω . RMS current: I = (V/X_C) = (110/177) ≈ 0.621 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²),

Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance

View Details

AC Through Inductor - Inductive Reactance

A \( 45 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is th

**Inductive reactance calculation** for 60 mH, 50 Hz, X_L=2π×50×0.06=18.85 Ω, V_rms=220 V, I_rms=220/18.85=11.67 A, for 70 mH, 50 Hz, X_L=21.99 Ω, I=110/21.99=5 A, showing X_L ∝ f L, higher f or L increases opposition. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 45 × 10⁻⁶ F . X_C = (1/376.8 × 45 × 10⁻⁶) ≈ 59 Ω . RMS current: I = (V/X_C) = (110/59) ≈ 1.864 A . Peak current: i_m = √(2) I = 1.414 × 1.864 ≈ 2.64 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²),

Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance

View Details

AC Through Inductor - Inductive Reactance

A \( 60 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the r

**Inductive reactance calculation** for 60 mH, 50 Hz, X_L=2π×50×0.06=18.85 Ω, V_rms=220 V, I_rms=220/18.85=11.67 A, for 70 mH, 50 Hz, X_L=21.99 Ω, I=110/21.99=5 A, showing X_L ∝ f L, higher f or L increases opposition. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 60 × 10⁻³ H . X_L = 314 × 0.06 = 18.84 Ω . RMS current: I = (V/X_L) = (220/18.84) ≈ 11.68 A . 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 11.68 A, consistent

Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance

View Details

Electrical Power, Energy and Heating Effect

A \( 10 \, \Omega \) resistor dissipates \( 40 \, \text{W} \) of power. What is the voltage across it?

**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. Power: P = (V²/R) . Rearrange: V = √(P R) . Substitute: V = √(40 × 10) = √(400) = 20 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 20 V,

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

View Details

Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 30 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three resistors \( 3 \, \Ome

**Kirchhoff's loop rule** (energy conservation) states algebraic sum of potential differences around closed loop zero, Σ ε - Σ I R =0, ensures total voltage drop across series resistors equals source emf. Junction rule (charge conservation) Σ I_in = Σ I_out ensures current continuity. Parallel resistance: (1/R_p) = (1/3) + (1/6) + (1/12) = (4 + 2 + 1/12) = (7/12) ⇒ R_p = (12/7) ≈ 1.71 Ω . Total resistance: Rtₒtₐl = 3 + 1.71 = 4.71 Ω . Total current: I = (ε/Rtₒtₐl) = (30/4.71) ≈ 6.37 A . Voltage across parallel: V = I R_p = 6.37 × 1.71 ≈ 10.89 V

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

View Details

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

View Details

EMF, Internal Resistance and Cells Combination

A \( 10 \, \text{V} \) battery with negligible internal resistance is connected to a \( 2 \, \Omega \) and \( 3 \, \Omeg

**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: R = 2 + 3 = 5 Ω . Current: I = (V/R) = (10/5) = 2 A . Power: P = I² R = 2² × 3 = 12 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

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

View Details