Question

An alternating current varying sinusoidally with a frequency of 50 Hz has an RMS value of 20 A. Write down the equation for the instantaneous value and find this value (a) 0.0025 second (b) 0.0125 second after passing through a positive maximum value. At what time, measured from a positive maximum value, will the instantaneous current be 14.14 A ?

Comments

How it changes from sin to cos ?

---

Time value are given from point A where voltage has positive and maximum values the equation may itself be referred to point A. In the case, the equation becomes
i = 28.2 cos 100πt
cos0° = 1
sin0° = 0

Answer 1

I_m = 20√2 = 28.2 A, ω = 2π × 50 = 100 π rad/s. 

The equation of the sinusoidal current wave with reference to point O (Fig.) as zero time point is 

i = 28.2 sin 100 πt ampere 

Since time values are given from point A where voltage has positive and maximum value, the equation may itself be referred to point A. In the case, the equation becomes : 

i = 28.2 cos 100 πt 

(i) When t = 0.0025 second

 i = 28.2 cos 100π × 0.0025 ...angle in radian 

= 28.2 cos 100 × 180 × 0.0025 ...angle in degrees 

= 28.2 cos 45º = 20 A ...point B

(ii) When t = 0.0125 second 

i = 28.2 cos 100 × 180 × 0.0125 

= 28.2 cos 225º = 28.2 × ( 1/ √2) − = −20 A ...point C 

(iii) Here i = 14.14 A 

∴ 14.14 = 28.2 cos 100 × 180 t 

∴ cos 100 × 180 t = 1/2 

or 100 × 180 t = cos⁻¹ (0.5) = 60º, t = 1/300 second ...point D