Question

An electric bulb has a rating of $500$W, $100$V. It is used in a circuit having a $200$V supply. What resistance must be connected in series with the bulb so that it delivers $500$ W?

• A. $10\Omega$
• B. $20\Omega$
• C. $30\Omega$
• D. $40\Omega$

Answer 1

The correct option is B $20\Omega$

$resistanceofbulb=p=\frac{v_n^2}{R^b}$

p is normal power

vn is nominal voltage

rb is resistance of bulb

$R_b=\frac{v_n^2}{p}=\frac{{100}^2}{500}$

nominal current $i_n=\frac{v_n}{R_b}=\frac{{100}^{}}{20}$

$=5A$

$R_{\Sigma }=\frac{v}{I_n}=\frac{200}{5}=40\Omega$

resistors equal to the sum

$R_{\Sigma }-R_b=40-20=20\Omega$