The coefficients of self induction of two coils are `L_(1)` and `L_(2)` .To induce an of `2` volt in the coil, a change of current of `1A` has to be produced in `5` second and `50ms` respectively. The ratio of their self inductances `L_(1):L_(2)` will be

The inductors of self inductance L_(1) and L_(2) are put in series .

The coefficients of self -induction of two coils are L_(1)=8mH and L_(2)=2mH respectively. The current rises in the two coils at the same rate . The power given to the two coils at any instant is same . The ratio of currents flowing in the coils will be

Coefficient of coupling between two coils of self-inductances L_(1) and L_(2) is unity. It means

Mutual inductance of two coils depends on their self inductance L_(1) and L_(2) as :

The self inductance of a coil is 5 henry , a current of 1 amp change to 2 amp within 5 second through the coil. The value of induced e.m.f. will be

The self induced emf of a coil is 25 volts. When the current in it is changed at uniformed rate from 10 A to 25 A in 1s, the change in the energy of the inductances is

If the rate of change of current is 1A per second in one coil induces an emf of 2 V in the neighbouring coil, the mutual inductance of two coils is

In Fig. coil 1 and coil 2 are wound on a long cylindrical insulator. The ends A' and B are joined together and current I is passed. Self-inductance of the two coils are L_(1) and L_(2) , and their mutual inductance is M . a. Show that this combination can be replaced by a single coil of equivalent inductange given by L_(eq) = L_(1) + L_(2) + 2M . b. How could the coils be reconnected by yieldings an equivalent inductance of L_(eq) = L_(1) + L_(2) - 2M .

Self-inductance of a coil is 50mH. A current of 1 A passing through the coil reduces to zero at steady rate in 0.1 s, the self-induced emf is