# In a step down transformer, how can low voltage give rise to large current, if V is proportional to I (by ohms law)? How is total energy conserved in a transformer?

### Asked by hari krishnan | 11th Nov, 2012, 06:37: PM

### A transformer is a device for stepping-up, or stepping-down, the voltage of an alternating electric signal. Without efficient transformers, the transmission and distribution of AC electric power over long distances would be impossible. Figure 51 shows the circuit diagram of a typical transformer. There are two circuits. Namely, the *primary circuit*, and the *secondary circuit*. There is no direct electrical connection between the two circuits, but each circuit contains a coil which links it *inductively* to the other circuit. In real transformers, the two coils are wound onto the same iron core. The purpose of the iron core is to channel the magnetic flux generated by the current flowing around the primary coil, so that as much of it as possible also links the secondary coil. The common magnetic flux linking the two coils is conventionally denoted in circuit diagrams by a number of parallel straight lines drawn between the coils.

Let us consider a particularly simple transformer in which the primary and secondary coils are *solenoids* sharing the same air-filled core. Suppose that is the length of the core, and is its cross-sectional area. Let be the total number of turns in the primary coil, and let be the total number of turns in the secondary coil. Suppose that an alternating voltage

is fed into the primary circuit from some external AC power source. Here, is the peak voltage in the primary circuit, and is the alternation frequency (in radians per second). The current driven around the primary circuit is written

where is the peak current. This current generates a changing magnetic flux, in the core of the solenoid, which links the secondary coil, and, thereby, inductively generates the alternating emf

in the secondary circuit, where is the peak voltage. Suppose that this emf drives an alternating current

around the secondary circuit, where is the peak current.
The circuit equation for the primary circuit is written

assuming that there is negligible resistance in this circuit. The first term in the above equation is the externally generated emf. The second term is the back-emf due to the self inductance of the primary coil. The final term is the emf due to the mutual inductance of the primary and secondary coils. In the absence of any significant resistance in the primary circuit, these three emfs must add up to zero. Equations 281, 282, 284 and285 can be combined to give

since

The alternating emf generated in the secondary circuit consists of the emf generated by the self inductance of the secondary coil, plus the emf generated by the mutual inductance of the primary and secondary coils. Thus,

Now, the instantaneous power output of the external AC power source which drives the primary circuit is

(290)

Likewise, the instantaneous electrical energy per unit time transfered inductively from the primary to the secondary circuit is

(291)

If resistive losses in the primary and secondary circuits are negligible, as is assumed to be the case, then, by energy conservation, these two powers must equal one another at all times. Thus,

(292)

which easily reduces to

Equations 286,289 and 293 yield

(294)

which gives

(295)

and, hence,

Equations 293 and 296 can be combined to give

(297)

Note that, although the mutual inductance of the two coils is entirely responsible for the transfer of energy between the primary and secondary circuits, it is the self inductances of the two coils which determine the ratio of the peak voltages and peak currents in these circuits.
the self inductances of the primary and secondary coils are given by and , respectively. It follows that

(298)

and, hence, that

*primary circuit*, and the

*secondary circuit*. There is no direct electrical connection between the two circuits, but each circuit contains a coil which links it

*inductively*to the other circuit. In real transformers, the two coils are wound onto the same iron core. The purpose of the iron core is to channel the magnetic flux generated by the current flowing around the primary coil, so that as much of it as possible also links the secondary coil. The common magnetic flux linking the two coils is conventionally denoted in circuit diagrams by a number of parallel straight lines drawn between the coils.

Let us consider a particularly simple transformer in which the primary and secondary coils are *solenoids* sharing the same air-filled core. Suppose that is the length of the core, and is its cross-sectional area. Let be the total number of turns in the primary coil, and let be the total number of turns in the secondary coil. Suppose that an alternating voltage

where is the peak current. This current generates a changing magnetic flux, in the core of the solenoid, which links the secondary coil, and, thereby, inductively generates the alternating emf

in the secondary circuit, where is the peak voltage. Suppose that this emf drives an alternating current

around the secondary circuit, where is the peak current.

The circuit equation for the primary circuit is written

assuming that there is negligible resistance in this circuit. The first term in the above equation is the externally generated emf. The second term is the back-emf due to the self inductance of the primary coil. The final term is the emf due to the mutual inductance of the primary and secondary coils. In the absence of any significant resistance in the primary circuit, these three emfs must add up to zero. Equations 281, 282, 284 and285 can be combined to givesince

The alternating emf generated in the secondary circuit consists of the emf generated by the self inductance of the secondary coil, plus the emf generated by the mutual inductance of the primary and secondary coils. Thus,

Now, the instantaneous power output of the external AC power source which drives the primary circuit is

(290) |

(291) |

(292) |

Equations 286,289 and 293 yield

(294) |

(295) |

Equations 293 and 296 can be combined to give

(297) |

the self inductances of the primary and secondary coils are given by and , respectively. It follows that

(298) |

### Answered by | 14th Nov, 2012, 11:19: AM

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