Question

8.5-1 The transformer described in the book (consisting of coils on opposite sides of a toroidal core) is modified as follows: A slot is cut through the core on the secondary side, and the secondary coil is rewound through the slot. As a result, the secondary coil remains oriented in the same direction relative to the magnetic flux, but has exactly one-half the cross-sectional area as it did previously (i.e., is now one-half the cross-sectional area of the primary coil). The overall cross-sectional dimensions of the core remains nearly the same on both sides, so the flux density in the core is not significantly affected. The number of turns in the primary and secondary coils are 200 and 300, respectively. What is the potential V2 on the secondary side in terms of the potential V1 on the primary side?

Answer 1

Given Information:

Number of primary turns = N₁ = 200

Number of secondary turns = N₂ = 300

Cross-sectional area of secondary = A₂ = ½A₁

Required Information:

Voltage on secondary side = V₂ = ?

Answer:

Voltage on secondary side = V₂ = 0.75V₁

Step-by-step explanation:

The emf induced in a transformer is given by

E = 4.44NfФ

Where N is the number of turns, f is the frequency and Ф is the magnetic flux and is equal to

Ф = BA

Where B is the magnetic field and A is the cross-sectional area.

The emf induced in the primary side of transformer is,

E₁ = 4.44N₁fBA₁

The emf induced in the secondary side of transformer is,

E₂ = 4.44N₂fBA₂

Since it is given that A₂ = ½A₁

E₂ = 4.44N₂fB½A₁

Then,

E₁/E₂ = (4.44N₁fBA₁)/(4.44N₂fB½A₁)

E₁/E₂ = (N₁)/(N₂½)

E₁/E₂ = (200)/(300*½)

E₁/E₂ = (200)/(150)

E₂ = 200/150E₁

E₂ = 0.75E₁

or

V₂ = 0.75V₁

Therefore, the secondary side voltage of the transformer is 0.75V₁