Question

A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V. Calculate the maximum instantaneous current through the filament.

Answer 1

Maximum Instantaneous Current is 0.39 A

Explanation:

• As per Ohm's Law,  

V = IR (where V is Voltage, I = Current and r = Resistance)

Also Power (P) =$I \times V\ or\ I^2 \times R\ or\ \frac {V^2}{R}$

P = 60W  

V = 220 V = E  

R = $\frac {V^2}{P}$  

= $\frac {(220 x 220)}{60}$

= 806.67

$\varepsilon _0 = \sqrt 2 E$

= $1.414 \times 220$

= 311.08

$I_0 = \frac {\varepsilon _0}{R}$

= $\frac {806.67}{311.08}$

= 0.385 $\approx$ 0.39 A