# Electrical Engineering - RL Circuits - Discussion

Discussion Forum : RL Circuits - General Questions (Q.No. 2)
2.
The voltages in Problem 5 are measured at a certain frequency. To make the resistor voltage less than the inductor voltage, the frequency is
increased
decreased
doubled
not a factor
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

For this the inductor resistance need to reduce and will done when frequency decreases.

ABHINAV KUMAR said:   1 decade ago
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.

XL= 2*3.14*f*L

Here inductive reactance increases by increasing frequency, so the voltage across the inductor increases.

Danish baidar said:   1 decade ago
To make the resistor voltage less than the inductor voltage.
Inductor voltage has to be increased.

VL=I(2*pi*f*L)

So f is Increased to increase Inductor voltage.

Ganesh shind said:   1 decade ago
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:   9 years ago
Generally is formula Xl=2piFL.

So. frequency is increased.

Brundha said:   9 years ago
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:   9 years ago
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:   8 years ago
But we have to inductive voltage more than resistance voltage so we have to decrease Xl.

Surendra said:   7 years ago
XL = 2 * &phi * f * L.

The frequency increase so the voltage across the inductor also increased.