A light bulb has the rating $200 W$ $220 V$
(1) Find the peak value of current
(2) If the supply voltage is $200 V$ , then find power consumed by the bulb

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Solution

Resistance $R = V^{2} / P = 48400 / 200 = 242 Ω$

$I_{r m s} = V_{r m s} / R = 220 / 242 = 0.909 A$

Peak Value of current $= I_{r m s} \times \sqrt[2]{2} = 0.909 \times 1.414 = 1.285 A$
$(2)$
Power $P = V_{1}^{2} / R = 200 \times 200 / 242 = 165.28 W$

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A light bulb has the rating 200W 220V(1) Find the peak value of current(2) If the supply voltage is 200V, then find power consumed by the bulb

Resistance R=V2/P=48400/200=242ΩIrms=Vrms/R=220/242=0.909APeak Value of current =Irms×2√2=0.909×1.414=1.285A (2)Power P=V21/R=200×200/242=165.28W

A light bulb has the rating $200 W$ $220 V$
(1) Find the peak value of current
(2) If the supply voltage is $200 V$ , then find power consumed by the bulb

Resistance $R = V^{2} / P = 48400 / 200 = 242 Ω$

$I_{r m s} = V_{r m s} / R = 220 / 242 = 0.909 A$

Peak Value of current $= I_{r m s} \times \sqrt[2]{2} = 0.909 \times 1.414 = 1.285 A$
$(2)$
Power $P = V_{1}^{2} / R = 200 \times 200 / 242 = 165.28 W$