IIT JEE Physics Formula Revision 39. Alternating current

Ch. 39 Alternating Current

1. Alternating current

i = i₀ sin (ωt + φ) .. (1)

where
i = current at time t
i₀ = peak current
current repeats after each time interval T = 2 π/ω

2. Emf produced in a AC generator

Є = NBA ω sin ωt = Є₀ sin ωt … (2)

Where
N = number of turns on the armature
B = Magnetic field
A = area of the coil on the armature
ω = angular velocity of the armature

3. RMS (rms) current in a AC circuit

i_rms = i₀/√2 … (3)
The alternating current and voltage are generally measured and expressed in termsof their rms values. When the household current is expressed as 220 V AC, it means that the rms value of voltage is 220 V. The peak voltage value is (220 V) √2 = 311 V.

4. current in an AC circuit containing only resistor

i = i₀ sin ωt …… (4)
where
i₀ = Є₀/R ... (5)

6. Ac circuit containing only a capacitor

i = i₀ cos ωt … (6)

where
i₀ =C Є₀ω
= Є₀/(1/ωC) … (7)

8. AC circuit containing only an inductor

i = (Є₀/ωL) sin (ωt – π/2) …(8)

or i = i₀ sin (ωt – π/2)
where i₀ = Є₀/ωL

10. Impedance

The peak current and the peak emf in Ac circuits may be written as

i₀ = Є₀/Z … (10)

where Z = R for a purely resistive circuit
Z = 1/ ωC for a purely capacitive circuit
Z = ωL for a purely inductive circuit

11. Vector method to find the current in an AC circuit

i = (Є₀/Z) sin (ωt + φ)

where
Z = √(R² + X²)
R = resistance in the circuit
X = reactance in the circuit (X due to capacitance is 1/ ωC and X due to inductance is ωL)

φ is to be determined from tan φ = X/R


15. Power in Ac Circuits

P = Є_rms i_rmscos φ

16. Voltages in primary and secondary of transformer

Є₂ = - (N₂ / N₁) Є₁ .... (16)

Power transfer

Є₁i₁ = Є₂ i₂ …(17)

i₂ = - (N₁ / N₂) i₁

‘-‘ sign shows that i₂ is 180° out of phase with i₁ .



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