Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 3)
3.
In a series RL circuit, 12 V rms is measured across the resistor, and 14 V rms is measured across the inductor. The peak value of the source voltage is
Discussion:
15 comments Page 2 of 2.
Pravin said:
1 decade ago
Consider,
Two voltage are connected in series so the two are addition so simple as this Que.
12+14 = 26.
Two voltage are connected in series so the two are addition so simple as this Que.
12+14 = 26.
D.P.MISHRA said:
1 decade ago
Sqrt(VL^2+VR^2)=Vrms.
i.e sqrt12^2+14^2=sqrt340.
Vp=sqrt2*Vrms.
i.e Sqrt(2*340)=sqrt680=26.0V approx.
i.e sqrt12^2+14^2=sqrt340.
Vp=sqrt2*Vrms.
i.e Sqrt(2*340)=sqrt680=26.0V approx.
Gerald said:
1 decade ago
The relationship between peak and rms voltages is given as: Vrms = (Vpeak)/sqrt(2).
As a continuation to Samuel Amu's solution, his answer (18.439V) is still the rms value.
Transforming to peak value gives:
Vpeak = (Vrms)*sqrt(2) = 18.439*sqrt(2) = 26.07 Volts.
As a continuation to Samuel Amu's solution, his answer (18.439V) is still the rms value.
Transforming to peak value gives:
Vpeak = (Vrms)*sqrt(2) = 18.439*sqrt(2) = 26.07 Volts.
Chris said:
1 decade ago
Current can't be constant, its a AC signal. If so, Z(L) would be zero.
But option B isn't correct either cause RMS isn't the same as peak.
But option B isn't correct either cause RMS isn't the same as peak.
Samuel amu said:
1 decade ago
In a series circuit, current is constant. assume a unity current flowing in the cct., then XL=VL/I, nb I=1(assumed). XL=14/1=14
and R=VR/I=12. cct impedance Z=(R^2+XL^2)^(-2)
Z=(12^2+14^2)^(-2)=18.439
peak voltage: Vp=IZ
Vp=1x18.439
Vp=18.439 v
So answer is Option A.
and R=VR/I=12. cct impedance Z=(R^2+XL^2)^(-2)
Z=(12^2+14^2)^(-2)=18.439
peak voltage: Vp=IZ
Vp=1x18.439
Vp=18.439 v
So answer is Option A.
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