Three impedances, each of (3−4j) Ω, are connected in delta to a 200 V, three-phase, 50 Hz balanced supply. What are the values of line current and phase current, respectively, in the delta-connected load?
$ 40 \sqrt3 A and 40 A$
To find the line current and phase current in a delta-connected load, we first need to understand the relationship between line current, phase current, and the impedance in a delta connection.
Step-by-Step Solution:
- The given impedance for each phase in the delta is (3 - 4j)\, \Omega.
- First, calculate the magnitude of the phase impedance: |Z| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, \Omega.
- Given the line voltage across the delta connection is 200 V, the phase voltage is the same as the line voltage in a delta connection. Therefore: V_{phase} = 200\, V.
- Calculate the phase current using Ohm's Law, I_{phase} = \frac{V_{phase}}{|Z|}: I_{phase} = \frac{200}{5} = 40\, A.
- In a delta-connected system, the line current I_{line} is related to the phase current I_{phase} by: I_{line} = \sqrt{3} \times I_{phase}.
- Therefore, the line current is: I_{line} = \sqrt{3} \times 40 = 40 \sqrt{3}\, A.
- Hence, the values of the line current and phase current in the delta-connected load are 40√3 A and 40 A, respectively.
Conclusion:
The correct answer is: 40 \sqrt{3} A (line current) and 40 A (phase current).
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