# Electronics - RL Circuits - Discussion

### Discussion :: RL Circuits - General Questions (Q.No.2)

2. Calculate the voltage dropped across R1 in the given circuit.

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

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?

 Vivek Yadav said: (Sep 12, 2020) What is the voltage drop across L1? Please explain anyone.