Question

An electric bell has a rating of 500 W and 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 $\Omega$

Given:
The bulb with power(P) 500 W operating at potential difference(V) 100 V.

We know that,
$P=I\times V$
Substituting the values, current(I) can be calucated as,
$⟹I=\frac{500}{100}=5A$

Resistance of bulb=
$R_2=\frac{V}{I}$$=\frac{100}{5}=20$ $\Omega$

To deliver 500 W the current in the bulb must remain 5 A when it is operated with 200 V supply. The resistance, $R_1$ to be connected in series for this purpose is given by:

$\frac{200}{R_1+R_2}=5$
$\Rightarrow$$R_1+R_2=40$
$\Rightarrow$$R_1+20=40$
$\Rightarrow$$R_1=20\Omega$.