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MCQEasyJEE 2025Alternating Current Basics

JEE Physics 2025 Question with Solution

An electric bulb rated as 100 W100 \, \text{W}-220 V220 \, \text{V} is connected to an ac source of rms voltage 220 V220 \, \text{V}. The peak value of current through the bulb is :

  • A

    0.64 A0.64 \, \text{A}

  • B

    0.45 A0.45 \, \text{A}

  • C

    2.2 A2.2 \, \text{A}

  • D

    0.32 A0.32 \, \text{A}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Power of the bulb is P=100 WP = 100 \, \text{W} and rms voltage is Vrms=220 VV_{\text{rms}} = 220 \, \text{V}.

Find: The peak current through the bulb.

For a resistive bulb,

P=VrmsIrmsP = V_{\text{rms}} I_{\text{rms}}

So,

Irms=PVrms=100220=511 A≈0.4545 AI_{\text{rms}} = \frac{P}{V_{\text{rms}}} = \frac{100}{220} = \frac{5}{11} \, \text{A} \approx 0.4545 \, \text{A}

The peak current is related to rms current by

I0=2 IrmsI_0 = \sqrt{2}\, I_{\text{rms}}

Substituting,

I0=2×511≈1.414×0.4545≈0.643 AI_0 = \sqrt{2} \times \frac{5}{11} \approx 1.414 \times 0.4545 \approx 0.643 \, \text{A}

Therefore, the peak value of current through the bulb is 0.64 A0.64 \, \text{A}. The correct option is A.

Using Resistance First

Given: P=100 WP = 100 \, \text{W}, Vrms=220 VV_{\text{rms}} = 220 \, \text{V}.

Find: Peak current through the bulb.

Using the power relation for a resistive bulb,

P=Vrms2RP = \frac{V_{\text{rms}}^2}{R}

Hence,

R=Vrms2P=2202100=484 ΩR = \frac{V_{\text{rms}}^2}{P} = \frac{220^2}{100} = 484 \, \Omega

Now find the rms current:

Irms=VrmsR=220484≈0.4545 AI_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{220}{484} \approx 0.4545 \, \text{A}

Then,

Ipeak=2 Irms≈1.414×0.4545≈0.643 AI_{\text{peak}} = \sqrt{2}\, I_{\text{rms}} \approx 1.414 \times 0.4545 \approx 0.643 \, \text{A}

Therefore, the peak current is 0.64 A0.64 \, \text{A}, so the correct option is A.

Common mistakes

  • Using P=VIP = VI with peak voltage or peak current instead of rms values is incorrect because the bulb rating is given in rms terms. First use VrmsV_{\text{rms}} and IrmsI_{\text{rms}}, then convert to peak current.

  • Taking Ipeak=Irms/2I_{\text{peak}} = I_{\text{rms}}/\sqrt{2} is wrong. For sinusoidal current, the correct relation is Ipeak=2 IrmsI_{\text{peak}} = \sqrt{2}\, I_{\text{rms}}.

  • Using the bulb rating 100 W−220 V100 \, \text{W}-220 \, \text{V} as two unrelated numbers can lead to confusion. The rating means the bulb consumes 100 W100 \, \text{W} when connected across 220 V220 \, \text{V} rms.

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