Question

A $50\text{W},100\text{V}$ lamp is to be connected to an $AC$ mains of $200\text{V},50\text{Hz}$. What capacitance is essential to be put in series with the lamp?

• A. $1\mu \text{F}$
• B. $2\mu \text{F}$
• C. $4.2\mu \text{F}$
• D. $9.2\mu \text{F}$

Answer 1

The correct option is D $9.2\mu \text{F}$

As resistance of the lamp

$R=\frac{V^2}{P}=\frac{{100}^2}{50}=200\Omega$

and the maximum current,

$i=\frac{V}{R}=\frac{100}{200}=0.5\text{A}$

So, when the lamp is put in series with a capacitance and connected to a $200\text{V}$ $AC$, using $V=iZ$, we get,

$Z=\frac{V}{i}=\frac{200}{0.5}=400\Omega$

Now, as in case of $CR$ circuit,

$Z=\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}$

$\Rightarrow R^2+{\left(\frac{1}{\omega C}\right)}^2=160000$

$\Rightarrow {\left(\frac{1}{\omega C}\right)}^2=16\times {10}^4-(200{)}^2=12\times {10}^4$

So, $\frac{1}{\omega C}=\sqrt{12}\times {10}^2$

$\Rightarrow C=\frac{1}{100\pi \times \sqrt{12}\times {10}^2}$

i.e., $C=\frac{{10}^{-4}}{\pi \sqrt{12}}=\frac{100\times {10}^{-6}}{\pi \sqrt{12}}\text{F}\approx 9.2\mu \text{F}$

Hence, option $\left(D\right)$ is correct.