Electronics - RL Circuits - Discussion

2. 

Calculate the voltage dropped across R1 in the given circuit.

[A]. 14 V
[B]. 26.8 V
[C]. 28 V
[D]. 0 V

Answer: Option B

Explanation:

No answer description available for this question.

Ria said: (Aug 17, 2011)  
Please explain.

Scott said: (Aug 26, 2011)  
First find XL for L1 => 2*pi*20 000*10x10^-6 = 1.256 ohms
after that find impedance which is sqr(R1^2+XL^2) = sqr(10^2+1.256^2) = 10.0786 ohms
and use voltage divider Vs(R1/R total) = 27(10/10.0786) = 26.789 V or 26.8 V

Sekhar said: (Jan 10, 2012)  
Please give me the explanation.

Adeel Awan said: (Jul 4, 2012)  
XL for L1 => 2*pi*20 000*10x10^-6 = 1.256 ohms

Sqr(R1^2+XL^2) = sqrroot(10^2+1.256^2) = 10.0786 ohms

And use voltage divider Vs(R1/R total) = 27(10/10.0786) = 26.789 V or 26.8 V

Amit Kumar Sahoo said: (Feb 20, 2014)  
Find REACTANCE XL using XL = 2*pi*f*L.
Find total impedance Z = (R^2+XL^2)^1/2.

Using voltage divider formula find voltage across R.
VOLTAGE ACROSS R = TOTAL VOLTAGE*(R/Z).

Dipanjan said: (Jan 31, 2015)  
XL1 = 2*pi*f*L=2*3.14*20*10^3*10^-5 = 1.256 ohms.

Z1 = sqrt[(R1^2)+(XL1)^2] = sqrt(10^2+1.256^2) = 10.0786 ohms.

Voltage dropped across R1 = Vs*(R1/Z1) = 27*(10/10.0786) = 26.8 V.

Prasanna Kumar said: (Apr 20, 2016)  
Voltage across resistor, R1= Resistance value * Supply voltage/Total resistance.
VR1 = R1*Vs/ (sqrt(R1^2+(2*pi*f*L)^2).
VR1 = 10*27/(sqrt(10^2 + (2*pi*20*10^3*10*10^- 6)^2).
= 26.78V.

Kiran V said: (Dec 15, 2016)  
I think voltage is same at any point in a parallel circuit.

Blank said: (Oct 16, 2017)  
FOR DC VOLTAGE inductor treated as a short circuit?

Saran said: (Apr 2, 2019)  
R1 is 10K, then how it will be R1=10^2?

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