Question

On a bulb is written $220V$ and $60W$. Find out the resistance of the bulb and the value of the current flowing through it:

• A. $500\Omega \text{and}2A$
• B. $200\Omega \text{and}4A$
• C. $100\Omega \text{and}1A$
• D. $806.67\Omega \text{and}0.27A$

Answer 1

Answer: (D)

Given,

Power rating of the bulb, $P=60W$

Power supply, $V=220V$

Resistance, $R$

Current, $I$

We know,

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

$⟹R=\frac{V^2}{P}$

$R=\frac{\left({220}^2\right)}{60}$

$R=806.67\Omega$

According to Ohm's law,

$V=IR$

$⟹I=\frac{V}{R}$

$I=\frac{220}{806.67}$

$I=0.27A$