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
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Prasad hajare said:   6 years ago
The frequency increase so the voltage across the inductor also increased. This one right conclusion.

Saubhagya said:   7 years ago
I can't get this question. Anyone explain me clearly.

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

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

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

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.

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.

Dhiva said:   10 years ago
Generally is formula Xl=2piFL.

So. frequency is increased.

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.

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.

Kishore said:   1 decade ago
XL= 2*3.14*f*L

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


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