## Friday, February 8, 2008

### 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

irms = 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 = Єrms irmscos 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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