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=50(rad)/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).
Which of the following graph roughly matches the variation of current in the circuit (with the coil and capacitor connected in the series) when the angulr frequency is decreased from 50 rad/s to 25 rad/s?

Options

A.
B.
C.
D.

Answer

B

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.