Electronics and Communication Engineering - Measurements and Instrumentation - Discussion

Q = XL/R.

Q=WL/R.

Here, q=(1/r*(l/c)^0.5) and r=kl/a.

R is decrease because of we know, R = rho*(L/A).

So area increase than are decrease, and area is increase because of thickness increase. So C is correct answer.

Q factor of a Coil is XL/R,

Q = ωL/R.

= 2If*L/R

So it is clear from above equation, that when are is will rise, Q will decrease. So, to increase Q factor, are should be decrease.

An ideal inductor would have no resistance or energy losses. However, real inductors have winding resistance from the metal wire forming the coils. Since the winding resistance appears as a resistance in series with the inductor, it is often called the series resistance. The inductor's series resistance converts electric current through the coils into heat, thus causing a loss of inductive quality.

The quality factor (or Q) of an inductor is the ratio of its inductive reactance to its resistance at a given frequency, and is a measure of its efficiency. The higher the Q factor of the inductor, the closer it approaches the behavior of an ideal, lossless, inductor. High Q inductors are used with capacitors to make resonant circuits in radio transmitters and receivers. The higher the Q is, the narrower the bandwidth of the resonant circuit.

The Q factor of an inductor can be found through the following formula, where L is the inductance, are is the inductor's effective series resistance, ω is the radian operating frequency, and the product ωL is the inductive reactance:.

Q = ωL/R.

Q = 2If*L/R. So, clearly if R is low, Q will increase.