A lamp having a resistance of $280 o h m$ is connected to a $E_{r m s} = 200 v o l t$ line. The peak value of the current passing through the lamp will be

4.0 A

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3.0 A

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2.0 A

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1.0 A

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Solution

The correct option is D $1.0 A$
Given,
$R = 282 Ω E_{R M S} = 200 V E_{M} = 200 \times \sqrt{2} = 282.84 V$
Now
$I = \frac{E_{m}}{R} = \frac{282.84}{280} = 1.01 A$
Correct option is D

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A lamp having a resistance of 280 ohm is connected to a Erms=200 volt line. The peak value of the current passing through the lamp will be

The correct option is D 1.0AGiven, R=282ΩERMS=200VEM=200×√2=282.84VNow I=EmR=282.84280=1.01ACorrect option is D

A lamp having a resistance of $280 o h m$ is connected to a $E_{r m s} = 200 v o l t$ line. The peak value of the current passing through the lamp will be

The correct option is D $1.0 A$
Given,
$R = 282 Ω E_{R M S} = 200 V E_{M} = 200 \times \sqrt{2} = 282.84 V$
Now
$I = \frac{E_{m}}{R} = \frac{282.84}{280} = 1.01 A$
Correct option is D