Two 220; V , 100 ;W bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220

Q: Two $220\; V , 100 \;W$ bulbs are connected first in series and then in parallel. Each time the combination is connected to a $220 \;V \;AC$ supply line. The power drawn by the combination in each case respectively will be

b $R=\frac{V^2}{P}=\frac{220 \times 220}{100}=484 \Omega$ In series, $R_{e q}=484+484+968 \Omega$ $\therefore P_{e q}=\frac{V^2}{968}=\frac{220 \times 220}{968}=50 \;watt$ In parallel, $R_{\text {eq }}=242 \Omega$ $\therefore P_{e q}=\frac{V^2}{242}=\frac{220 \times 220}{242}=200\; watt$.

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Two $220\; V , 100 \;W$ bulbs are connected first in series and then in parallel. Each time the combination is connected to a $220 \;V \;AC$ supply line. The power drawn by the combination in each case respectively will be

A$50\; watt, 100 \;watt$
B$50 \;watt, 200 \;watt$
C$100\; watt, 50 \;watt$
D$200 \;watt, 150\; watt$

AIPMT 2003, Medium

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

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