Current Electricity 4 Question 14
14. An electric bulb rated for $500 W$ at $100 V$ is used in a circuit having a $200 V$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb delivers $500 W$ is …… $\Omega$.
$(1987,2 M)$
Analytical & Descriptive Question
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Solution:
In parallel, current distributes in inverse ratio of resistance. Hence,
$$ \frac{I-I _g}{I _g}=\frac{G}{S} \Rightarrow S=\frac{G I _g}{I-I _g} $$
As $I _g$ is very small, hence
$$ \begin{aligned} S & =\frac{G I _g}{I} \\ b & =\frac{(100)\left(1 \times 10^{-3}\right)}{10}=0.01 \Omega \end{aligned} $$
