# Shunt Resistance of Cavity Resonator

In the given diagram L and C is the equivalent ckt of the cavity resonator. Here L represents the walls of the cavity and C represent the mouth of the cavity. If we use the resistance in parallel to the tuned ckt (Cavity Resonator), this will cause to decrease the total resistance of the cavity resonator. The power loses will decrease because

P = I^{2}R

The Q of the cavity will increase, without changing the volume etc of the cavity, here the power output of the cavity will increase.

Problem:

The total impedance on the cavity resonator is 10 MW Ω. What should be the volume of the shunt resistance to decrease the power losses by 50 %?

Z = 10 MW

P = I^{2}R

.

Suppose

I^{2} = I amp

Therefore

P = 1 X 10^{6} W

Now to get 50% power loss we are to used a shunt resistor equals to value of the impedance of the cavity so value of shunt resistor should be equal to 10 MW?

Prove:-