Question

An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It

पर्याय

• must be zero

• may be zero

• is never zero

• `is (220//sqrt2 )V`

Answer 1

may be zero

Let the AC voltage be given by,
`V = V_0sin  omegat`
Here, ω = `2pi` f = 314 rad/s
The average voltage over the given time,
Here, ω = `2pi` f = 314 rad/s
The average voltage over the given time,
Vavg = \[\frac{\int_0^{0.01}Vdt}{\int_0^{0.01}dt}\]=` -V_0 [(cos omega)/omega]_0^0.01`
= `V_0/(wxx0.01) (1 - cospi)`

= `(2V_0)/pi=140.127V` 
Also, when V = V_0 cos `omegat`
Vavg =\[\frac{\int_0^{0.01}Vdt}{\int_0^{0.01}dt}\]=` -V_0 [(cos omega)/omega]_0^0.01`

=` V_0 /(omegaxx0.01) (sin omega(0.01)-0)`

=`V_0/(314xx0.01)(sin omega (0.01)-0)`

= `V_0/3.14 (sinpi)`
=0

From the above results, we can say that the average voltage can be zero. But it is not necessary that it should be zero or never zero.