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An `AC` voltage is applied acrss a series combination of `L` and `R`. If the voltage drop across the resistor and inductor are `20 V` and `15 V` respectiely, then applied peak voltage isA. `25V`B. `35V`C. `25sqrt2V`D. `5sqrt7V` |
Answer» Correct Answer - C `V=sqrt(V_R^2+V_L^2)` `=sqrt((20)^2(15)^2)=25V` But this is the rms value. `:.` Peak value of `=sqrt2 V_("rms")=25sqrt(2)V`. |
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