Question 2List three functions which a transformer may perform in an electrical circuit. [3 marks]A 10kVA, 50 Hz distribution transformer has 600 turns in the primary and 120 turns in thesecondary. The primary and secondary resistances are r1 = 0.2 Ω and r 2 = 0.02 Ω respectively,and the corresponding leakage reactances are Xl1 = 1.0 Ω, and Xl2 = 0.04 Ω respectively. Thesupply voltage is 1100V.Calculate(a) The equivalent impedance, resistance and reactance, referring all quantities to thehigh voltage (primary) side; [6 marks](b) The secondary terminal voltage for full load having a power factor of 0.7 lagging;Draw the phasor diagram; [7 marks](c) The regulation in (b) [4 marks]2130310310
Question
Question 2List three functions which a transformer may perform in an electrical circuit. [3 marks]A 10kVA, 50 Hz distribution transformer has 600 turns in the primary and 120 turns in thesecondary. The primary and secondary resistances are r1 = 0.2 Ω and r 2 = 0.02 Ω respectively,and the corresponding leakage reactances are Xl1 = 1.0 Ω, and Xl2 = 0.04 Ω respectively. Thesupply voltage is 1100V.Calculate(a) The equivalent impedance, resistance and reactance, referring all quantities to thehigh voltage (primary) side; 6 marks The secondary terminal voltage for full load having a power factor of 0.7 lagging;Draw the phasor diagram; 7 marks The regulation in (b) [4 marks]2130310310
Solution
Sure, here are the answers to your questions:
(a) The equivalent impedance, resistance and reactance, referring all quantities to the high voltage (primary) side:
First, we need to find the turns ratio (a) of the transformer, which is the ratio of the number of turns in the primary coil to the number of turns in the secondary coil.
a = N1/N2 = 600/120 = 5
Next, we can find the equivalent resistance (Re) and reactance (Xe) on the primary side by adding the primary resistance/reactance to the secondary resistance/reactance multiplied by the square of the turns ratio.
Re = r1 + (a^2 * r2) = 0.2 + (5^2 * 0.02) = 0.2 + 5 = 5.2 Ω
Xe = Xl1 + (a^2 * Xl2) = 1.0 + (5^2 * 0.04) = 1.0 + 10 = 11.0 Ω
The equivalent impedance (Ze) is the square root of the sum of the squares of the equivalent resistance and reactance.
Ze = sqrt(Re^2 + Xe^2) = sqrt(5.2^2 + 11.0^2) = sqrt(27.04 + 121) = sqrt(148.04) = 12.17 Ω
(b) The secondary terminal voltage for full load having a power factor of 0.7 lagging:
The secondary terminal voltage (V2) can be found using the formula:
V2 = V1 - (I2 * Ze * cosθ)
where V1 is the primary voltage, I2 is the secondary current, Ze is the equivalent impedance, and cosθ is the power factor.
First, we need to find the secondary current (I2), which is the power (P) divided by the product of the secondary voltage (V2') and the power factor (cosθ).
I2 = P / (V2' * cosθ) = 10000 / (1100/5 * 0.7) = 8.06 A
Then, we can find the secondary terminal voltage:
V2 = V1 - (I2 * Ze * cosθ) = 1100 - (8.06 * 12.17 * 0.7) = 1100 - 68.6 = 1031.4 V
(c) The regulation in (b):
The regulation (R) is the difference between the no-load secondary voltage and the full-load secondary voltage, divided by the full-load secondary voltage, expressed as a percentage.
R = ((V1 - V2) / V2) * 100% = ((1100 - 1031.4) / 1031.4) * 100% = 6.65%
Please note that the phasor diagram is a graphical representation and cannot be drawn in text.
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