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When an ac source of emf`e=E_(0) sin (100 t)` is connected across a circuit, the phase difference between emf e and currnet I in the circuit is observed to be `(pi)//(4)` as shown in fig. If the circuit consists possibly only of R-C or R-C of L-R series, find the relationship find the relationship between the two elements. A. `R = 1 k Omega, C = 10 mu F`B. `R = 1 k Omega, C = 1 mu F`C. `R = 1 k Omega, L = 10 mH`D. `R = 10 k Omega, L = 10 mu H` |
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Answer» Correct Answer - A Fig. shows that current `i` leads the emf `e` by a phase angle `pi//4`. Therefore, the circuit can be `R - C` circuit alone. `tan phi = (X_(C ))/(R ) = (1)/(omega CR) = "tan" (pi)/(4) = 1` or `CR = (1)/(omega)` From `e = E_(0) sin 100 t, omega = 100 rad//s` `:. CR = (1)/(omega) = (1)/(100)` when `R = 1 K Omega = 10^(3) Omega` `C = (1)/(omega R) = (1)/(10^(5)) = 10^(-5) F = 10 mu F` |
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