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A resistance of `40 Omega` is connected to an source of 220 v, 50 Hz. Find the (i) rms current (ii) maximum instantaneous current in resistor. |
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Answer» Here, `R = 40 Omegam E_(v) = 220 V, v = 50 Hz`, (i) `I_(v) = (E_(v))/( R) = (220)/(40) = 5.5 A` `I_(v) = sqrt 2 I_(v) = 1.414 xx 5.5 = 7.78 A` If alternating current is given by `I = I_(0) sin omega t`, then `I_(0) = I_(0) sin omega t_(1) = 1` or `omega t_(1) = (pi)/(2)` and `I_(v) = (I_(0))/(sqrt 2) = I_(0) sin omega t_(2)`, which implies `omega t_(2) (pi)/(2) + (pi)/(4)` `:. omega (t_(2) - t_(1)) = (pi)/(2) + (pi)/(4) - (pi)/(2) = (pi)/(4)` `t_(2) - t_(1) = (pi)/(4 omega) = (pi)/(4 xx 2 pi v) = (1)/(8 v) = (1)/(8 xx 50)` `= 2.5 xx 10^(-3) s = 2.5 ms` |
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