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When an AC source of `emf e=E_(0) sin(100t)` is connected across a circuit i in the circuit, the phase difference between the emf e and the current i in the circuit is observed to be `(pi//4)`, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C in series, find the relationship between the two elements A. `R=1k Omega, C=10 mu F`B. `R=1k Omega, C=1 mu F`C. `R=1k Omega, L=10 mu F`D. `R=1k Omega, L=1 mu F` |
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Answer» Correct Answer - A Since,current leads emf (as seen from the graph), therefore, this is an R-c circuit. `tan(phi)=(X_(C)_X_(L))/(R)` Here, `(phi)=(45^@)` `:. (X_C)=R` [`(X_L)=0` as there is no inductor] `1/(omega C)=R implies RC(omega)=1` `:. RC=(1)/(100)(s^(-1))`. |
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