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प्रश्न
An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It
विकल्प
must be zero
may be zero
is never zero
`is (220//sqrt2 )V`
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उत्तर
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.
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