A symmetrical three-phase 240 v supplies a balanced delta connected load 6-8j ohms per phase. What is the line current?
Question
A symmetrical three-phase 240 v supplies a balanced delta connected load 6-8j ohms per phase. What is the line current?
Solution
To find the line current in a balanced delta connected load, we can follow these steps:
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First, we need to find the phase voltage. In a delta connected system, the line voltage is equal to the phase voltage. So, the phase voltage (Vph) is 240V.
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Next, we calculate the phase current. The phase current (Iph) is the phase voltage (Vph) divided by the phase impedance (Zph). The phase impedance given is 6-8j ohms. So, Iph = Vph / Zph = 240 / (6-8j) = 20A with a phase angle of 53.13 degrees.
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In a delta connected system, the line current (IL) is √3 times the phase current (Iph) and leads it by 30 degrees. So, IL = √3 * Iph = √3 * 20 = 34.64A. The phase angle of the line current is 53.13 + 30 = 83.13 degrees.
So, the line current is 34.64A with a phase angle of 83.13 degrees.
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