Question

A single-phase transformer has a turns-ratio of 144/432 and
operates at a maximum flux of 7.5 mWb at 50 Hz. When
working on no load, it takes 0.24 kVA at a power factor of 0.26
lagging from the supply. If it supplies a load of 1.2 kVA at a
power factor of 0.8 lagging, determine (a) the magnetizing
current, (b) the primary current, and (c) the primary power
factor.

Answer 1

Given that,

Turns ratio = 144/432

Maximum flux = 7.5 mWb

Frequency = 50 Hz

We need to calculate the primary voltage

Using formula of voltage

$V_{p}=4.44\times f\times\phi\times N_{p}$

Put the value into the formula

$V_{p}=4.44\times50\times7.5\times10^{-3}\times144$

$V_{p}=239.76\ volt$

We need to calculate the secondary voltage

Using formula of voltage

$V_{s}=4.44\times f\times\phi\times N_{s}$

Put the value into the formula

$V_{s}=4.44\times50\times7.5\times10^{-3}\times432$

$V_{s}=719.28\ volt$

At no load,

We need to calculate the value of $\phi$

Using formula of factor

$\text{power factor}=\cos\phi$

$\phi=\cos^{-1}{\text{power factor}}$

Put the value into the formula

$\phi=\cos^{-1}(0.26)$

$\phi=74.92^{\circ}$

We need to calculate the current $I_{0}$

Using formula of power

$P=V_{p}\times I_{0}$

$I_{0}=\dfrac{P}{V_{p}}$

Put the value into the formula

$I_{0}=\dfrac{0.24\times10^{3}}{239.76}$

$I_{0}=1.001\ A$

(I). We need to calculate the magnetizing current

Using formula of magnetizing

$I_{m}=I_{0}\sin\phi$

Put the value into the formula

$I_{m}=1.001\times\sin(74.96)$

$I_{m}=0.9667\ A$

Now, no load current is

$I_{0}=1.001<-74.92\ A$

At load,

We need to calculate the value of $\phi$

Using formula of factor

$\text{power factor}=\cos\phi$

$\phi=\cos^{-1}{\text{power factor}}$

Put the value into the formula

$\phi=\cos^{-1}(0.8)$

$\phi=36.87^{\circ}$

We need to calculate the secondary current

Using formula of power

$P=V_{s}\times I_{s}$

$I_{s}=\dfrac{P}{V_{s}}$

Put the value into the formula

$I_{s}=\dfrac{1.2\times10^{3}}{719.28}$

$I_{s}=1.668\ A$

Now, load current is

$I_{0}=1.668<-36.87\ A$

(II). We need to calculate the primary current

Using formula of primary current

$I_{p}=\dfrac{N_{s}}{N_{p}}\times I_{s}$

Put the value into the formula

$I_{p}=\dfrac{432}{144}\times 1.668$

$I_{p}=5.004<-36.87\ A$

We need to calculate the total current

Using formula of total current

$I=I_{p}+I_{0}$

Put the value into the formula

$I=(5.004<-36.87)+(1.001<-74.92)$

$I=5.825<-42.95\ A$

(III). We need calculate the power factor

Using formula power factor

$\text{power factor}=\cos\phi$

$\text{power factor}=\cos(42.95)$

$\text{power factor}=0.7319$

Hence, (I). The magnetizing current is 0.9667 A.

(II). The primary current is $5.004<-36.87\ A$

(III). The primary power factor is 0.7319.