Electrical Engineering - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 20)
20.
Five inductors are connected in series. The lowest value is 8 μH. If the value of each inductor is twice that of the preceding one, and if the inductors are connected in order of ascending values, the total inductance is
Discussion:
4 comments Page 1 of 1.
NT- said:
1 decade ago
Or you could realize that you have a geometric series with 5 terms and use the formula.
s = a* (1-r^n) / (1-r) where in this case a is 8 uH, n is 5, and R is 2.
s = a* (1-r^n) / (1-r) where in this case a is 8 uH, n is 5, and R is 2.
(1)
Prasanthi rao said:
1 decade ago
The inductances values are 8,16,32,64,128
Therefore total inductance connected in series = 8+16+32+64+128 = 248
Therefore total inductance connected in series = 8+16+32+64+128 = 248
D.RAMESH said:
9 years ago
Total inductance = L1 + L2 + L3 + L4 + L5.
L1 = 8 L2 = 16 L3 = 32 L4 = 64 L5 = 128.
8 + 16 + 32 + 64 + 128 = 248.
L1 = 8 L2 = 16 L3 = 32 L4 = 64 L5 = 128.
8 + 16 + 32 + 64 + 128 = 248.
Srikanth said:
8 years ago
Yes, correct @D.Ramesh.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers