Question
Download Solution PDFA circuit element 'X' when connected to peak voltage of 200 V, a peak current of 5A flows which lags behind the voltage by
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
Let Voltage V = VmSin(wt)
V = 200Sin(wt)
When connected to "X",
The current flowing lags "V" by
I = 5Sin(wt -
Since the current lags "V" by
Then reactance "X" would be = 200/5 = 40 ohm
The reactance "X" = 40j.
When connected to "Y" , the current 5 A is in phase with the voltage.
So, the "Y" will be a resistor.
Resistance "Y" = 200/5 = 40 ohm.
The "X" and "Y" are connected in series to the same applied voltage.
The net impedance would be = 40 + 40j
The current I' flowing, when connected in series, will be,
Peak Value of I' =
The RMS value of I' will be =
= 5/2
= 2.5 A.
The correct answer is option (3)
Last updated on May 15, 2025
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