Electronics - RL Circuits - Discussion
Discussion Forum : RL Circuits - Filling the Blanks (Q.No. 11)
11.
The voltage dropped across the resistor in the circuit in the given circuit is approximately equal to ___.
Discussion:
11 comments Page 1 of 2.
Unicorn said:
4 years ago
Vltg across a R=Vs √(R/Z).
Vs = 10, R = 1k ,Z = √(R^2 +XL^2).
Z=1252.59.
The answer will be Vtg across R=7.98.
Vs = 10, R = 1k ,Z = √(R^2 +XL^2).
Z=1252.59.
The answer will be Vtg across R=7.98.
Pritam said:
5 years ago
Inductor is a storing element then why there is a voltage drop?
Ajit dubey said:
9 years ago
Voltage divider have a been done between two resistor not unlike impedance.
Rushi said:
9 years ago
Why it can't be solved by using VTG divider?
PrivUsr said:
9 years ago
Why can't I use voltage divider in this case.
Such that, Vs = Vr * [R1/(R1+RL)].
Such that, Vs = Vr * [R1/(R1+RL)].
Gauri said:
10 years ago
Is there in any shortcut method to avoid all such tedious calculations.
Ravendra kushwaha said:
1 decade ago
I = V/Z.
Z = ROOT(R^2 + XL^2).
XL = 2Pi*F*L.
XL = 2*3.143*5000*.024.
XL = 754.32.
Z = ROOT(1000^2 + 754.32^2 ).
Z = ROOT(1000000+ 568998.662).
Z = ROOT(1568998.662).
Z = 1252.59.
Vs = 10v.
Z = 1252.59Ohm.
I = 10/1252.59.
I = 0.00798Amp.
Voltage dropped across the resistor R1, R1 = 1000Ohm.
V = I*R1.
V = 0.00798*1000.
V = 7.983volt.
Z = ROOT(R^2 + XL^2).
XL = 2Pi*F*L.
XL = 2*3.143*5000*.024.
XL = 754.32.
Z = ROOT(1000^2 + 754.32^2 ).
Z = ROOT(1000000+ 568998.662).
Z = ROOT(1568998.662).
Z = 1252.59.
Vs = 10v.
Z = 1252.59Ohm.
I = 10/1252.59.
I = 0.00798Amp.
Voltage dropped across the resistor R1, R1 = 1000Ohm.
V = I*R1.
V = 0.00798*1000.
V = 7.983volt.
R.Mageshwari said:
1 decade ago
Small correction is there on the answer clarification.
Voltage V=IR not VR.
Voltage V=IR not VR.
Clara said:
1 decade ago
Where Z=sqrt((R^2)+(XL^2))
XL=2*Pi*F*L ==> 2*3.14*5K*24m ==> 753.6
VR=I*R
substituting the values:
Z=sqrt((1k^2)+(753.6)^2) ==>1252.16
I=10/1252.16 ==>7.98mA
Finally VR=7.98m*1K ==>7.98V
XL=2*Pi*F*L ==> 2*3.14*5K*24m ==> 753.6
VR=I*R
substituting the values:
Z=sqrt((1k^2)+(753.6)^2) ==>1252.16
I=10/1252.16 ==>7.98mA
Finally VR=7.98m*1K ==>7.98V
Clara said:
1 decade ago
I=V/Z
where Z=sqrt((R^2)+(XL^2))
XL=2*Pi*F*L ==> 2*3.14*5K*24m ==> 753.6
VR=I*R
substituting the values:
Z=sqrt((1k^2)+(753.6)^2) ==>1252.16
I=10/1252.16 ==>7.98mA
Finally VR=7.98m*1K ==>7.98V
where Z=sqrt((R^2)+(XL^2))
XL=2*Pi*F*L ==> 2*3.14*5K*24m ==> 753.6
VR=I*R
substituting the values:
Z=sqrt((1k^2)+(753.6)^2) ==>1252.16
I=10/1252.16 ==>7.98mA
Finally VR=7.98m*1K ==>7.98V
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