A `10 mu F` capacitor is charged to a potential of 25 V. The battery is then disconnected and a pure 100 mH coil is connected across the capacitor so that LC oscillations are set-up. What is the maximum current in the coil ?

Answer

`0.25`

Explanation

Here, capacitance of the capacitor,
`C = 10muF = 10xx10^(-6)F=10^(-5)F`
Inductance of the coil, `L = 100mH = 10^(-1)H`
Voltage to which the capacitor is charged, i.e., `V_0 = 25 V`
Electric energy of the capacitor `=1/2 CV_0^2`
Since LC oscillations are set up, that means there is continuous conversion of electric energy into magnetic energy and vice-versa. Further, as the inductance is pure, no energy is wasted in the . entire electric energy is change i.e., `1/2 LI_0^2` where `I_0` , is maximum current in the coil,
i.e., `1/2 LI_0^2 =1/2 CV_0^2`
or `I_0=V_0sqrt(C/L)=25sqrt((10^(-5)F)/(10^(-1)H))`
`=25 xx10^(-2)A =0.25A`