A single-phase transformer has 400 primary and 1000 secondary turns. T

1.	A single-phase transformer has 400 primary and 1000 secondary turns. The net crosssectional area of the core is 60 . If the primary winding be connected to a 50 Hz supply at 500 V, calculate i) the peak value of the flux density in the core, and ii) the voltage induced in the secondary winding
Answer» Given: Primary turns, n1 = 400 Secondary turns, n2 =1000 Cross sectional area of the core = 60 sq.cm. Frequency, f = 50 Hz Voltage APPLIED, V = 500 V (i) Peak value of FLUX density at core, $B_{max}$ $B_{max}=\frac{V}{4.44 fNA}$ where N is the no. of turns on the side voltage is applied and f is in MHz. Substituting the given values in the formula, we have $B_{max}=\frac{500}{4.44 5010^{-6}40060}=0.9384 Tesla$ (ii) The voltage induced in secondary side is determined by turns ratio and primary voltage. We have the relationship, $\frac{No. of turns on primary side}{No. of turns on secondary side}= \frac{Voltage on primary side}{Voltage on secondary side}$ Now, Voltage on secondary side = (N2/N1) * Primary voltage =(1000/400) * 500 = 1250 V Therefore, the voltage induced in the secondary side of the transformer is 1250 volts.

A single-phase transformer has 400 primary and 1000 secondary turns. The net crosssectional area of the core is 60 . If the primary winding be connected to a 50 Hz supply at 500 V, calculate i) the peak value of the flux density in the core, and ii) the voltage induced in the secondary winding

Answer»

Given:

Primary turns, n1 = 400

Secondary turns, n2 =1000

Cross sectional area of the core = 60 sq.cm.

Frequency, f = 50 Hz

Voltage APPLIED, V = 500 V

(i) Peak value of FLUX density at core, $B_{max}$

$B_{max}=\frac{V}{4.44 fNA}$

where N is the no. of turns on the side voltage is applied

and f is in MHz.

Substituting the given values in the formula, we have

$B_{max}=\frac{500}{4.44 *50*10^{-6}*400*60}=0.9384 Tesla$

(ii) The voltage induced in secondary side is determined by turns ratio and primary voltage.

We have the relationship,

$\frac{No. of turns on primary side}{No. of turns on secondary side}= \frac{Voltage on primary side}{Voltage on secondary side}$

Now, Voltage on secondary side = (N2/N1) * Primary voltage

=(1000/400) * 500

= 1250 V

Therefore, the voltage induced in the secondary side of the transformer is 1250 volts.