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An alternative voltage given by V = 140 sin 314t is connected across a pure resistor of 50 Ω. Find The frequency of the source. The rms current is through the resistor.

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प्रश्न

An alternative voltage given by V = 140 sin 314t is connected across a pure resistor of 50 Ω. Find

  1. The frequency of the source.
  2. The rms current is through the resistor.
योग
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उत्तर

Given:

V = 140 sin 314t, R = 50Ω

  1. Comparing with V = V0 sinωt
    Thus, V0 = 140 V
    ω = 314
    2πν = 314
    `"ν" = 314/(2 xx3.14) "Hz" = 50" Hz"`
  2. `"V"_("rms") = "V"_0/sqrt2 = 140/sqrt2 = 98.99"V"`
    `"I"_("rms") = "V"_("rms")/"R" = 98.99/50 = 1.97"A"`
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2011-2012 (March) All India Set 1

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