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

Answer: Option B

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.

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