1.

An alternating voltage V = 400 sin (500omegat)V is connected to a 0.2 kOmegaresistor, what will be the rms value of current ?

Answer»

14.14 A
1.414 A
0.1414 A
`2A`

Solution :Comparing EQUATION V=400 SIN `(500omegat)` with `V=V_m sin OMEGAT`,
`V_m`=400 V `therefore V_"RMS"=V_m/sqrt2`
`I_"rms"=V_"rms"/R=V_m /(sqrt2R)=400/(sqrt2xx200)`
`=2/sqrt2=sqrt2`=1.414 A


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