Question

The operating temperature of the filament of lamp is $2000{}^{\circ }C$. The temperature coefficient of the material of filament is $0.005{}^{\circ }{C}^{-1}$. If the atmospheric temperature is $0{}^{\circ }C$, then the current in the $100W-200V$ lamp when it is switched on is nearest to

• A. $2.5A$
• B. $3.5A$
• C. $4.5A$
• D. $5.5A$

Answer 1

The correct option is D $5.5A$

$\alpha =0.005{}^{\circ }{C}^{-1}$
Power = $100W$
We know that $P=\frac{V^2}{R_t}$
$⟹100=\frac{(200{)}^2}{R_t}$
$⟹R_t=400\Omega$

The change in resistance with temperature is given by
$R_t=R_0[1+\alpha (\Delta T\left)\right]$
(where $R_0$ is the resistance at $0{}^{\circ }C$)
$400=R_0[1+0.005(2000-0\left)\right]$
$R_0=\frac{400}{11}\Omega$
Now, we know that $I=\frac{V}{R_0}$
$⟹I=\frac{200}{\frac{400}{11}}$
$\therefore I=5.5A$