Question

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 power consumed by the bulb.

Answer 1

1.

The resistance is given as,

$R=\frac{V^2}{P}$

$R=\frac{{\left(220\right)}^2}{200}$

$R=242\Omega$

The rms current is given as,

$i_{rms}=\frac{v_{rms}}{R}$

$i_{rms}=\frac{220}{242}$

$i_{rms}=0.909$

The peak current is given as,

$i_p=i_{rms}\times \sqrt{2}$

$i_p=0.909\times \sqrt{2}$

$i_p=1.2856A$

2.

The new power consumed by the bulb is given as,

$P_1=\frac{V^2}{R}$

$P_1=\frac{{\left(200\right)}^2}{242}$

$P_1=165.29W$