A coil of inductance 150 mh is connected in series with a variable capacitor of capacitance 20 pf to 500 pf. Calculate the frequency range over which the circuit can be tuned.

4

explanation:

4

the answer is kevlar

answer(20) psi;

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j