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An electric lamp is connected to 220V, 50 Hz supply. Then the peak value of voltage is
  • a)
    210 V
  • b)
    211 V
  • c)
    310 V
  • d)
    320 V
Correct answer is option 'C'. Can you explain this answer?
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Understanding the Problem

To determine the peak value of voltage in an electric lamp connected to a 220 V, 50 Hz supply.

Explanation

When an electric lamp is connected to a supply, it operates at the same frequency as the supply. In this case, the supply frequency is given as 50 Hz.

Peak Value of Voltage

The peak value of voltage is the maximum value that the voltage reaches during each cycle of the alternating current. It is given by the formula:

Peak Voltage = RMS Voltage * √2

Where RMS voltage is the root mean square voltage, which is given by:

RMS Voltage = (Peak Voltage) / √2

Since the RMS voltage is given as 220 V, we can calculate the peak voltage using the above formula.

Calculating the Peak Voltage

RMS Voltage = (Peak Voltage) / √2

Rearranging the formula, we get:

Peak Voltage = RMS Voltage * √2

Substituting the given RMS voltage of 220 V into the formula:

Peak Voltage = 220 V * √2

Using a calculator, we can calculate the value of √2 as approximately 1.414.

Peak Voltage = 220 V * 1.414

Peak Voltage ≈ 310 V

Therefore, the peak value of voltage in the electric lamp connected to a 220 V, 50 Hz supply is approximately 310 V.

Answer

The correct answer is option C) 310 V.
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