Electrical Engineering - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 7)
7.
Two 10 H inductors are in parallel and the parallel combination is in series with a third 10 H inductor. What is the approximate total reactance when a voltage with a frequency of 7 kHz is applied across the circuit terminals?
219 k
66 k
660 k
1.3 M
Discussion:
13 comments Page 1 of 2.
Piyush mishra said:
5 years ago
XL = 2ΠfL = 2*3.14*7000*15 = 659400= That means Approximately 660k ohm.
(1)
Raja said:
1 decade ago
Can any one explain this why ?
Asha said:
1 decade ago
L equivalent=15H
reactance=wL=2*PIE*F*L
reactance=wL=2*PIE*F*L
Shruthi mathew said:
1 decade ago
1/L(equa) = 1/L1 + 1/L2.
L(equa) = 5H.
L(equa)+10H = 15H.
XL = wL
= 2*3.14*f*L
= 2*3.14*7*1000*15
= 659.4*1000
= 660kohms.
L(equa) = 5H.
L(equa)+10H = 15H.
XL = wL
= 2*3.14*f*L
= 2*3.14*7*1000*15
= 659.4*1000
= 660kohms.
Naresh said:
1 decade ago
XL = wL.
= 2*3.14*f*L.
= 2*3.14*7*1000*15.
= 659.4*1000.
= 660kohms.
= 2*3.14*f*L.
= 2*3.14*7*1000*15.
= 659.4*1000.
= 660kohms.
P.Ramachandran said:
1 decade ago
1/L=1/L1+1/L2+L3
=1/10+1/10+10
L=1/0.2+10
L= 5+10
L=15H
XL=2*3.14*F*L
=2*3.14*7*15
=659.4KOHM
=1/10+1/10+10
L=1/0.2+10
L= 5+10
L=15H
XL=2*3.14*F*L
=2*3.14*7*15
=659.4KOHM
KISHAN TJ said:
1 decade ago
XL=2*PIE*F*L
=2*3.14*7*1000*15
=6.28*7*1000*15
=659.4*1000
=660Kohms
=2*3.14*7*1000*15
=6.28*7*1000*15
=659.4*1000
=660Kohms
Girisha P N said:
1 decade ago
XL = 2*Pi*f*L.
L = 15H.
XL = 0.6 Mohm.
L = 15H.
XL = 0.6 Mohm.
Kiran said:
9 years ago
why here 1000 is multiplied with 659.4 ohms?
Puja said:
9 years ago
I don't understand it. Someone explain me in detail.
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