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A `100 Omega` resistance is connected in series with a `4 H` inductor. The voltage across the resistor is, `V_(R ) = (2.0 V) sin (10^(3) t)`. Find amplitude of the voltage across the inductor.A. `40 V`B. `60 V`C. `80 V`D. `90 V` |
Answer» Correct Answer - C Amplitude of voltage across inductor, `V_(0) = I_(00 X_(L)` `= (2.0 xx 10^(-2) A) (4.0 xx 10^(3)` ohm) = 80 volts |
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