Question

An electric bulb of $500\text{W}$ at $100\text{V}$ is used in a circuit having a $200\text{V}$ supply. Calculate the resistance $R$ to be connected in series with the bulb so that the power delivered by the bulb is $500\text{W}$.

• A. $30\Omega$
• B. $5\Omega$
• C. $10\Omega$
• D. $20\Omega$

Answer 1

The correct option is D $20\Omega$

The resistance of the bulb is given by :
$R_B=\frac{V^2}{P}=\frac{{100}^2}{500}=20\Omega$

Let the current flowing through the bulb is $I$ as shown below :
According to the question :
$I^{2}R=500$
$I^2=\frac{500}{20}$
$I=5\text{A}$
Using KVL in the circuit :
$200=I(R_B+R)$
$40=20+R$
$\Rightarrow R=20\Omega$