#electrical power

Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit containing only a resistor, what happens to the power dissipated if the frequency of the source is doub

**Wattless current** occurs in pure inductor or capacitor, I_rms non-zero but average power zero because φ=±90°, cos φ=0, energy oscillates between source and field, no dissipation, used in choke coil to limit current without heating, unlike resistor where power dissipated. In a purely resistive AC circuit, power dissipated is P = I² R , where I = (V/R) , and R is constant. Since resistance does not depend on frequency, and assuming the rms voltage remains constant, the power dissipated remains unchanged when frequency doubles. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V

Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current

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AC Through Capacitor - Capacitive Reactance

A \( 130 \, \text{V} \) (rms) AC source supplies a \( 65 \, \Omega \) resistor. What is the average power consumed?

**AC through capacitor** voltage lags current by 90°, Q= C V, I= dQ/dt = C dV/dt, V(t)=V_peak sin ωt, I(t)=I_peak sin(ωt+90°), average power zero because energy stored in electric field ½ C V² returned each cycle, capacitor blocks DC but passes AC, X_C decreases with f. RMS current: I = (V/R) = (130/65) = 2 A . Average power: P = I² R = 2² × 65 = 260 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 260 W, consistent with phasor analysis and resonance condition X_L = X_C.

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

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Potentiometer, Conductivity and Special Cases

A \( 12 \, \text{V} \) battery with negligible internal resistance is connected to a \( 4 \, \Omega \) and \( 8 \, \Omeg

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Total resistance: R = 4 + 8 = 12 Ω . Current: I = (V/R) = (12/12) = 1 A . Power: P = I² R = 1² × 4 = 4 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

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

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Electrical Power, Energy and Heating Effect

A \( 8 \, \Omega \) resistor carries a current of \( 2 \, \text{A} \) for \( 15 \, \text{s} \). What is the energy dissi

**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. Energy: W = I² R t . Substitute: W = 2² × 8 × 15 = 4 × 120 = 480 J . 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 480 J,

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

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Electrical Power, Energy and Heating Effect

A \( 10 \, \Omega \) resistor dissipates \( 25 \, \text{W} \) of power. What is the current through it?

**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). Power: P = I² R . Rearrange: I = √((P/R)) . Substitute: I = √((25/10)) = √(2.5) ≈ 1.58 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 1.58 A,

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

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Electrical Power, Energy and Heating Effect

A circuit has a \( 10 \, \text{V} \) battery with negligible internal resistance connected to two resistors \( 5 \, \Ome

**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: R = 5 + 10 = 15 Ω . Current: I = (V/R) = (10/15) = (2/3) A . Power in 10 Ω : P = I² R = ((2/3))² × 10 = (4/9) × 10 = 4.44 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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Electrical Power, Energy and Heating Effect

A \( 24 \, \text{V} \) battery with negligible internal resistance is connected to a \( 4 \, \Omega \) and \( 8 \, \Omeg

**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: R = 4 + 8 = 12 Ω . Current: I = (V/R) = (24/12) = 2 A . Power: P = I² R = 2² × 4 = 16 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 16 W,

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

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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

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