Self Induction
Trending Questions
A wire of fixed length is wound on a solenoid of length and radius . Its Self Inductance is found to be . Now, if the same wire is wound on a solenoid of length and radius , then self inductance will be
- 16
- 12
- 8
- 4
- Zero
- 2 mH
- 2 H
- 200 mH
The equivalent quantity of mass in electricity is
Charge
Potential
Inductance
Current
Two different coils have self-inductance L1 = 8mH and L2 = 2mH respectively. The current in both the coils is increased at a same rate. At a certain instant of time, the power being supplied to both the coils is equal. At that instant, the current, the induced voltage, and the energy stored in L1 and L2 are I1 and I2, V1 and V2, and W1 and W2 respectively. Then
i1i2=14
i1i2=4 8
W2W1=4
V2V1=14
The coefficient of self inductance of a solenoid is 0.18 mH. If a rod of soft iron of relative permeability 900 is inserted in it, then the coefficient of self inductance will become nearly
5.4 mH
162 mH
0.006 mH
0.0002 mH
- 1 H
- 4 H
- 10−3 H
- 4× 10−3 H
- 100 V
- 10 V
- 1000 V
- 90 V
- Zero
- 0.5 H
- 50 H
- 5 mH
- 0.2H
- 2mH
- 2H
- 0.2mH
- 5 V
- 5×10−3V
- 0.5 V
- 5×10−2V
- 2 mH
- 3 mH
- 4 mH
- 5 mH
- Nine times
- Remain constant
- Halved
- Doubled
- 16
- 12
- 8
- 4
- Zero
- 0.5 H
- 50 H
- 5 mH