Electrical Engineering - Inductors - Discussion
Discussion Forum : 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
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 isDiscussion:
8 comments Page 1 of 1.
Coolman said:
6 years ago
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.
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.
(2)
Parashu nilogal said:
8 years ago
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.
=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.
(1)
Vijay rank said:
8 years ago
Input=output + losses.
=4*10^3 + 800.
=4*10^3 + 800.
Shuchi Jaiswal said:
8 years ago
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.
Please give me the answer.
Satish lali said:
9 years ago
Nice, Thank you @P.Ramachandran.
P.Ramachandran said:
1 decade ago
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
=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
Amit Thapar univ said:
1 decade ago
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
XL=2*Pi*f*L=2*3.14*300*1000*36.6*(1/1000000 )=68(approx)
current=V/XL=60/68=.87=870mA
Pradeep said:
1 decade ago
Can someone explain this?
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