Answer

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**Hint:**Get the relation as per the conditions given in the question between the peak power and the power consumed. This shall help in getting the phase difference. Power factor is a measure of how effectively you are using electricity. The ideal power factor is unity, or one. Anything less than one means that extra power is required to achieve the actual task at hand.

**Complete answer:**

Power is the amount of energy transferred per unit time.

The formula of power is given by:

$P = (V_{rms})(I_{rms})cos(φ)$

Where, φ is the phase difference between V and I, and the formula of Vrms and Irms is given by:

$

{V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }} \\

{I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }} \\

$

By substituting the values of Vrms and Irms in the formula of P, we get,

$P = \dfrac{{{V_0}{I_0}\cos \varphi }}{2}$

Therefore,

P = (P

_{peak})cos(φ),

where Ppeak is the maximum power that can be applied in an AC circuit.

Now, it is mentioned in the question that the lamp consumed only 50% of the maximum power applied in the AC circuit, which means,

\[\;\;P{\text{ }} = \dfrac{{{P_{peak}}}}{2}\]

By equating both the equations of power obtained, we get,

\[\;\;P{\text{ }} = \dfrac{{{P_{peak}}}}{2} = {P_{peak}}\cos \varphi \]

So, we get,

cos(φ) = ½

That is why the value of φ will be obtained as:

$\varphi = \dfrac{\pi }{3}rad$

So, the value of phase difference between the applied voltage and circuit current is \[\dfrac{\pi }{3}\] rad.

**So, the correct answer is “Option A”.**

**Note:**

Whenever we get this type of question the key concept of solving is that either we can proceed as done in this question or One can even use the phase diagrams to calculate the phase difference. The formula based approach as done here is also useful.

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