A student in a lab took a coil and connected it to a `12 VDC` source. He measures the steady state current in the circuit to be `4 A`. He then replaced the `12 VDC` source by a `12V,(omega=50rad//s)AC` source and observes that the reading in the `AC` ammeter is `2.4 A`. He then decides to connect a `2500 muF` capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).
The value of resistance of the coil calculated by the student is

Options

A. `3Omega`
B. `4Omega`
C. `5Omega`
D. `8Omega`

Answer

A

Explanation

`V_(DC)=I_(DC)R`
`:. R=V_(DC)/I_(DC)=12/4=3Omega`
`I_(AC)=V_(AC)/Z=V_(AC)/sqrt(R^2+X_L^(2))`
`2.4=12/sqrt((3)^2+X_L^(2))`
solving this equation we get,
`X_L=4Omega`
`X_C=1/(omegaC)=1/(50xx2500xx10^-6)`
`=8Omega`
`Z=sqrt(R^2+(X_C-X_L)^(2))=5Omega`
`:. I=V_(DC)/Z=12/5=2.4A=I_("rms")`
`P=I_("rms")^2R=(2.4)^2(3)`
`=17.28W`
At given frequency `X_CgtX_L` if omega is further decreased `X_C` will increas (as `X_Cprop1/omega`) and `X_L` will increase (as `X_Lprop omega`).
Therefore `X_C-X_L` and hence `Z` will increase. So, current will decrease.