When two inductors having self inductances L

_{1 }_{ }and L_{2 }_{ }are coupled in series, mutual inductance M exists between them. Two kinds of series connection are possible as follows : In this connection, two coils are connected in series such that their induced fluxes or voltages are additive in nature.

Fig. 1 |

Here currents i

_{1 }and i_{2 }is nothing but current i which is entering dots for both the coils. Self induced voltage in coil 1 = v

_{1 }_{ }= -L_{1 }_{ }(di/dt) Self induced voltage in coil 2 = v

_{2 }_{ }= -L_{2 }_{ }(di/dt) Mutually induced voltage in coil 1 due to change in current in coil 2 = v

_{1}**'**= -M(di/dt) Mutually induced voltage in coil 2 due to change in current in coil 1 = v

_{2}**'**= -M(di/dt) Total induced voltage = v

_{1 }_{ }+ v_{2 }_{ }+ v_{1}**'**+ v_{2}**'** = (L

_{1 }_{ }(di/dt) + L_{2 }_{ }(di/dt) + M(di/dt) + M(di/dt)) = - (L

_{1 }_{ }+ L_{2 }_{ }+ 2M)(di/dt) If L is equivalent inductance across terminals a-b then total induced voltage in single inductance would be equal to -L

_{eff }_{ }(di/dt). Comparing two voltages, L

Self induced voltage in coil 1 = -L

Self induced voltage in coil 2 = -L

Mutually induced voltage in coil 1 due to change in current in coil 2 = v

Also mutually induced voltage in coil 2 due to change in current in coil 1 = v

Therefore total induced voltage = v

= -L

= -(L

_{eff }_{ }= L_{1 }_{ }+ L_{2 }_{ }+ 2M__1.1 Series Opposing__

In this connection, two coils are connected in such a way that, the induced fluxes or voltages are of opposite polarities.

Here iFig. 2 |

_{1 }and i_{2 }is same series current ''i'' which is entering dot for coil L_{1 }_{ }and leaving dot for coil L_{2}.Self induced voltage in coil 1 = -L

_{1 }_{ }(di/dt)Self induced voltage in coil 2 = -L

_{2 }_{ }(di/dt)Mutually induced voltage in coil 1 due to change in current in coil 2 = v

_{1}**'**+ M (di/dt)Also mutually induced voltage in coil 2 due to change in current in coil 1 = v

_{2}**'**+M(di/dt)Therefore total induced voltage = v

_{1}+ v_{2}**+ v**_{1}**'**+ v_{2}**'**= -L

_{1 }_{ }(di/dt) - L_{2 }_{ }(di/dt) + M(di/dt) + (di/dt)= -(L

_{1 }_{ }+ L_{2 }_{ }- 2M)(di/dt) If L is equivalent inductance across terminals a and b then total induced voltage in single inductance would be equal to -L

L_{eff }(di/dt)). Comparing two voltages,_{eff }_{ }= L_{1 }_{ }+ L_{2 }_{ }- 2M

**Sponsored links :**
## 0 comments:

## Post a Comment