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KCET · Physics · Alternating Current

A light bulb rated \(100\) W is connected to an AC source of \(220\) V, \(50\) Hz. The rms current through the bulb is

  1. A \(0.454\) A
  2. B \(0.545\) A
  3. C \(2.20\) A
  4. D \(0.22\) A
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.454\) A

Step-by-step Solution

Detailed explanation

The power of the light bulb is given as \(P = 100\) W.

The rms voltage of the AC source is \(V_{rms} = 220\) V.

For a purely resistive load like a light bulb, the power factor is \(1\). The average power is given by the relation:

\(P = V_{rms} I_{rms}\)

Rearranging the formula to solve for the rms current \(I_{rms}\):

\(I_{rms} = \dfrac{P}{V_{rms}}\)

Substituting the given values:

\(I_{rms} = \dfrac{100}{220}\)

\(I_{rms} = \dfrac{5}{11} \approx 0.454\) A

Answer: \(0.454\) A