Question

A 100 kva, 2400 v/ 240 v, 50 hz single phase transformer has an exciting current of 0.64 a and core loss 700 w when its high voltage side is energized at rated voltage and frequency. If load current is 40 a at 0.8 pf lagging on the lv side, then magnitude of the primary current will be

Answer 1

Thus the magnitude of primary current is 0.57 A

Explanation:

• Primary voltage = 2400 V
• Exciting or no load current Io = 0.64 A
• Iron loss = 700 W

Iron loss current = I = Io cos ϕo  = Iron loss / V1 = 700 / 2400

Iron loss current = I 0.292 A

Magnetising component of no load current

Im = √ Io^2 - Ie^2 = √(0.64)^2 - (0.292)^2 = 0.57 A

Ro = V1 / Ie = 2400 / 0.292 = 8,222 Ω = 8.22 KΩ

Xo = V1 / Im = 2400 / 0.57 = 4120 Ω = 4.12 KΩ

Thus the magnitude of primary current is 0.57 A

Answer 2

Given that,

Frequency = 50 Hz

Primary voltage = 2400 V

Secondary voltage = 240 V

Exciting current = 0.64 A

Power = 700 W

Load current = 40 A

Lagging = 0.8 pf

We need to calculate the core loss current

Using formula of current

$I_{c}=\dfrac{P_{c}}{V_{1}}$

Where, P=power

V = voltage

Put the value into the formula

$I_{c}=\dfrac{700}{2400}$

$I_{c}=0.291\ A$

We need to calculate the magnetization current

Using formula of magnetization current

$I_{m}=\sqrt{I_{e}^2-I_{c}^2}$

Put the value into the formula

$I_{m}=\sqrt{0.64^2-0.291^2}$

$I_{m}=0.57\ A$

We need to calculate the phase angle

Using formula of phase angle

$\cos\theta=\dfrac{I_{c}}{I_{e}}$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{0.291}{0.64}$

$\theta=62.9^{\circ}$

We need to calculate the magnitude of the primary current

Using formula of current

$I_{1}=\dfrac{V_{2}}{V_{1}}\times I_{l}$

Where, $V_{2}$ = secondary voltage

$V_{1}$ = primary voltage

$I_{l}$ = load current

Put the value into the formula

$I_{1}=\dfrac{240}{2400}\times40$

$I_{1}=4\ A$

Hence, The magnitude of the primary current is 4 A.