Question

An electric bulb rated as $500W - 100V$ is used in a circuit having $200V$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb draws $500W$ is
(A) $100\Omega$
(B) $50\Omega$
(C) $20\Omega$
(D) $10\Omega$

Answer 1

The rating of the bulb tells you that the bulb consumes a particular amount of power when a particular voltage is applied across it. In this case, it consumes $500W$ when a source of $100V$ is applied across the bulb. First find the resistance of the bulb using the given rating of the bulb, then consider the case of $200V$ supply, find current flowing through the bulb and then use it to find the resistance needed to be added in series.

Formulae to be used: $P = \dfrac{{{V^2}}}{R}$
$P = {I^2}R$

Complete step by step solution:
The power consumed by the bulb is given as $P = \dfrac{{{V^2}}}{R}$
$R = \dfrac{{{V^2}}}{P}$
$R = \dfrac{{{{\left( {100} \right)}^2}}}{{500}} \\\ R = 20\Omega \\\$

The resistance of the bulb is $20\Omega$.

Now, consider the case when $200V$ is applied.

The current flowing through the bulb is given from$P = {I^2}R$

$$P = {I^2}R \\\ \therefore I = \sqrt {\dfrac{P}{R}} \\\$$

$
\therefore I = \sqrt {\dfrac{{500}}{{20}}} \\
\therefore I = 5A \\

$

The current flowing through the bulb will be $5A$.

Now, let the resistance to be added in series with the bulb be $R$.
The current which will flow through the bulb, the same amount of current will flow through the circuit.

Applying Ohm’s Law,
$V = I{R_{eq}}$, where ${R_{eq}}$ is the net resistance in the circuit.
Having $V = 200V$and $I = 5A$, resistance of the circuit will be given as ${R_{eq}} = \dfrac{V}{I}$
$\therefore {R_{eq}} = \dfrac{{200}}{5} \\\ \therefore {R_{eq}} = 40\Omega \\\$
From $40\Omega$, $20\Omega$ resistance is of the bulb. Therefore, the remaining $20\Omega$
is the contribution of the added resistance in series.

Therefore, the resistance $R$ that must be put in series with the bulb, so that the bulb draws
$500W$ is $20\Omega$.

Hence, Option (C) is correct.

Note: Remember that the resistance can be found using the rating of the bulb. In the equation$P = \dfrac{{{V^2}}}{R}$, do not use the voltage applied by the source to find the resistance, here the voltage $V$ is the voltage from the rating of the bulb.