Saved Bookmarks
| 1. |
A 200 mH (pure) inductor and a 5uF (pure) capacitor are connected, one by one, across a sinusoidal a.c. voltage source V= 70.7 sin (1000 t) volts. Obtain the expression for the current in each case. |
|
Answer» Solution :Here L = 200 mH = 0.2 H, `C = 5muF xx 10^(-6)F` and V = 70.7 sin(1000 t) means that `omega = 1000 s^(-1)` For inductive cirucit`I_(m) = V_(m)/X_(L) = V_(m)/(Lomega) = (70.7)/(0.2 xx 1000) = 0.354 A` So expression for current will be, `I_(L) = I_(m) sin (omega t -pi/2) = 0.354 sin(1000 t - pi/2) A` For CAPACITIVE circuit, `I_(m) = V_(m)/X_(L) =V_(m).C.omega = 70.7 xx 5 xx 10^(-6) xx 1000 = 0.354 A` `therefore` Expression for current will be, `I_( c) = I_(m) sin (omega t + pi/2) = 0.354 sin (1000 t + pi/2) A` |
|