Question

A single phase transformer has 500 primary and 1200 secondary turns. The net cross-sectional area of the core is 80cm². If the primary winding is connected to 60Hz supply at 750V. Calculate
I. The peak value of the flux linkage in the core
ii. The voltage induced in the secondary winding

Answer 1

To find the peak value of the flux linkage in the core, we need to find the magnetic flux in the core and multiply it by the number of turns in the primary winding.

The magnetic flux in the core can be found using the formula:

Φ = V t / (4.44  f A)

where V is the voltage applied to the primary winding, t is the time interval, f is the frequency of the supply, and A is the cross-sectional area of the core.

The peak value of the magnetic flux is equal to the maximum value of the voltage applied to the primary winding divided by 4.44 times the frequency of the supply times the cross-sectional area of the core:

Φ_peak = V_peak / (4.44 ×  f ×  A)

=$750 / (4.44 * 60 * 80 * 10^-4)$

= $0.0208 Wb$

The peak value of the flux linkage in the core is then:

λ_peak = N_p × Φ_peak

= $500 * 0.0208$

= $10.4 mWb$

II. To find the voltage induced in the secondary winding, we can use the formula:

V_secondary = N_secondary × λ_peak / t

where N_secondary is the number of turns in the secondary winding and t is the time interval.

Since the time interval is the same for both windings, we can write:

V_secondary = N_secondary × λ_peak / N_p × t

= $1200 * 10.4 / 500 * t$

= $25.92 t$

The voltage induced in the secondary winding is 25.92 times the time interval.

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