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The output E of an alternating voltage s...

The output E of an alternating voltage supply of frequency 50 Hz is shown in Fig. From this Fig, state
(i) value of time `t_(1)`
(ii) peak voltage `E_(0)` (iii) root mean square voltage `E_(v)`
(iv) mean supply is connected in series with a resistance of `2.4 Omega`, calculate the mean power dissipated in the resistor.

Text Solution

Verified by Experts

Here, `v = 50 Hz, E_(0) = 10 V`
From Fig. we find that
`t_(1) = (3)/(2) T = (3)/(2 xx v) = (3)/(2 xx 50) = 0.03 s`
(ii) peak voltage = 10 V
(iii) `E_(v) = (E_(0))/(sqrt2) = (10)/(1.414) = 7.07` volt
(iv) Mean averge voltage over full cycle = Zero. the mean power dissipated in the resistor
`E_(v) = (E_(v)^(2))/(R ) = (10 sqrt(2)^(2))/(2.4) = (50)/(2.4) = 20.83 W`
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