# Solved Problems-11

Problems-11

**A coil resonates at 2 MHz when a 18 pF capacitor is shunted across it. When shunting capacitor is 81 pF, the resonating frequency becomes 1 MHz. Find the distributed capacitor of the coil and the self resonating frequency.**

Solution

Let the distributed capacitance of the coil be

*C*(in pF)._{d}Show when the coil is shunted with another capacitance of

*C*, the total capacitance becomes,(

*C*+*C*)._{d}When

*C*= 18 pF, the resonating frequency is*f*_{0}= 2 MHz(i)

When

*C*= 81 pF, the resonating frequency is*f*_{0}= 1 MHz(ii)

Dividing (

*i*) by (*ii*), we get,Putting the value of

*C*_{d}in (*ii*), we get,So, the self-resonating frequency of the coil is given as,