Discussion :: Measurements and Instrumentation - Section 1 (Q.No.1)
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To increase Q factor of a coil, the wire should be
Answer: Option C
Q is high if resistance is low. Therefore, wire should be thick.
|Aurobinda said: (Nov 26, 2014)|
|Q = 2If*L/R. So, clearly if R is low, Q will increase.|
|Shivaraj said: (Feb 9, 2015)|
|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.
|Md Amir Rahman said: (Sep 11, 2015)|
|Q factor of a Coil is XL/R,
Q = ωL/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.
|Manish Patidar said: (Jan 9, 2016)|
|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.
|Rjv said: (Jul 31, 2017)|
|Here, q=(1/r*(l/c)^0.5) and r=kl/a.|
|Rishikesh Meena said: (Jan 27, 2018)|
|Shamee Kumar Singh said: (Jun 20, 2019)|
|Q = XL/R.|
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