Question

A filament bulb $\left(500\text{W}100\text{V}\right)$ is to be wed in a $230\text{V}$ main supply. When a resistance $R$ is connected in series, it works perfectly and the bulb consumes $500\text{W}$. The value $R$ is

• A. $26\Omega$
• B. $13\Omega$
• C. $230\Omega$
• D. $46\Omega$

Answer 1

The correct option is A $26\Omega$

Resistance of the bulb,
$R^1=\frac{V^2}{P}$ = $\frac{100\times 100}{500}$ = $20\Omega$
It is connected in series with $R$, equivalent resistance = $20+R$
Power consumed by the bulb,
$P=I^2\times R^1$
$\Rightarrow 500={\left(\frac{230}{20+R}\right)}^2\times 20$
$\Rightarrow (20+R{)}^2=(23{)}^2\times 4$
$\Rightarrow 20+R=23\times 2=46$
$\therefore R=26\Omega$