# Electrical Engineering - Inductors - Discussion

### Discussion :: Inductors - General Questions (Q.No.9)

9.

Two inductors, L1 and L2, are in parallel. L1 has a value of 25 H and L2 a value of 50 H. The parallel combination is in series with L3, a 20 H coil. The entire combination is across an ac source of 60 Vrms at 300 kHz. The total rms current is

 [A]. 87 mA [B]. 870 mA [C]. 87 A [D]. 337 mA

Explanation:

No answer description available for this question.

 Pradeep said: (Sep 17, 2011) Can someone explain this?

 Amit Thapar Univ said: (Sep 27, 2011) Combination will be L=(25*50/75)+20=36.6 XL=2*Pi*f*L=2*3.14*300*1000*36.6*(1/1000000 )=68(approx) current=V/XL=60/68=.87=870mA

 P.Ramachandran said: (Aug 24, 2012) 1/L=1/L1+1/L2 =1/25+1/50 =0.04+0.02 1/L=0.06 L=1/0.06 L=16.66 L=L12+L3 L=16.66+20 L=36.66(micro henry) L=0.0000366H F=300KHZ F=300*1000 F=30000 XL=2*3.14*300000*0.0000366 XL=68.95OHM I(RMS)=V/XL =60/68.95 =0.882A =882mA

 Satish Lali said: (Nov 2, 2016) Nice, Thank you @P.Ramachandran.

 Shuchi Jaiswal said: (Apr 11, 2017) The electrical energy required to heat a bucket of water to a certain temperature is 4kwh. If the heat losses are 20percentage, The energy input is? Please give me the answer.

 Vijay Rank said: (Jul 18, 2017) Input=output + losses. =4*10^3 + 800.

 Parashu Nilogal said: (Sep 21, 2017) 1/L=1/L1+1/L2 =1/25+1/50 =0.04+0.02 1/L=0.06 L=1/0.06 L=16.66 L=L12+L3 L=16.66+20 L=36.66(micro henry) XL=2*3.14*F*L XL=2*3.14*0.3*10^6*36.66*10^-6 XL=6.28*.3*36.6 XL=68.9 OHM V=60 I=V/XL I=60/68.9 I=.870 A I=870mA.

 Coolman said: (Oct 23, 2019) 1. L1 parallel with L2, which is ( 25 micro * 50 micro)/( 25 micro + 50 micro) = 16.67 micro Henry. 2. L1 and L2 parallel value series with L3 which is ( 16.67 Micro + 20 Micro) = 36.67 micro Henry. So now put into the formula, I rms= Vrms / 2*pai *F*L. = 60/( 2*pi*300K*36.67 micro). =0.87 A. = 870 mA.