A bulb rated at (100,W - 200,V) is used on a 100,V line. The current in the bulb is

Q: A bulb rated at ($100\,W$ - $200\,V$) is used on a $100\,V$ line. The current in the bulb is

a (a) $P = \frac{{{V^2}}}{R} \Rightarrow 100 = \frac{{{{(200)}^2}}}{R}$ $ \Rightarrow $ $R = \frac{{4 \times {{10}^4}}}{{{{10}^2}}} = 400\,\Omega $ Now, $i = \frac{V}{R} = \frac{{100}}{{400}} = \frac{1}{4}\,amp$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A bulb rated at ($100\,W$ - $200\,V$) is used on a $100\,V$ line. The current in the bulb is

A$\frac{1}{4}$ $amp$
B$4\, amp$
C$\frac{1}{2}$ $amp$
D$2\, amp$

Medium

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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