Discussion :: RL Circuits - General Questions (Q.No.2)
|Mani said: (Jun 23, 2011)|
|For this the inductor resistance need to reduce and will done when frequency decreases.|
|Abhinav Kumar said: (Nov 13, 2011)|
|In RL circuit both R & L are connected in series. Voltage drop is more at high impedance as V proportional to R whil I constant. So impedance of inductor should be high. Hence frequency should be increases as Zl=wL.|
|Kishore said: (Jan 23, 2012)|
Here inductive reactance increases by increasing frequency, so the voltage across the inductor increases.
|Danish Baidar said: (Apr 16, 2013)|
|To make the resistor voltage less than the inductor voltage.
Inductor voltage has to be increased.
So f is Increased to increase Inductor voltage.
|Ganesh Shind said: (Jul 20, 2013)|
|For resistive circuit voltage and current are in phase but for.
R-L circuit current lags the voltage by 90 degree so there is phase difference phi because of that in R-L circuit current is out of phase.
|Dhiva said: (May 28, 2014)|
|Generally is formula Xl=2piFL.
So. frequency is increased.
|Brundha said: (Jan 19, 2015)|
|To make the resistor voltage less than the inductor voltage?
Normally we know the formula for frequency,
xl = 2*3.14*l*f.
So, the voltage across the inductor is increased.
|Shardul said: (Mar 8, 2015)|
|Xl=2*3.14*f*l and Xc=1/2*3.14*f*c. So for in case of indicator its proportional form.|
|Sanu Singh said: (Dec 7, 2015)|
|But we have to inductive voltage more than resistance voltage so we have to decrease Xl.|
|Surendra said: (Sep 9, 2016)|
|XL = 2 * &phi * f * L.
The frequency increase so the voltage across the inductor also increased.
|Saubhagya said: (Jan 9, 2017)|
|I can't get this question. Anyone explain me clearly.|
|Prasad Hajare said: (Sep 7, 2017)|
|The frequency increase so the voltage across the inductor also increased. This one right conclusion.|
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