A heater coil connected to a supply of a 220, V is dissipating some power {P_1}. The coil is cut into half and the two

Q: A heater coil connected to a supply of a $220\, V$ is dissipating some power ${P_1}.$ The coil is cut into half and the two halves are connected in parallel. The heater now dissipates a power ${P_2}.$ The ratio of power ${P_1}\,\,:\,\,{P_2}$ is

c (c) $P = \frac{{{V^2}}}{R}$. If resistance of heater coil is $R$, then resistance of parallel combination of two halves will be $\frac{R}{4}$ So $\frac{{{P_1}}}{{{P_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{R/4}}{R} = \frac{1}{4}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A heater coil connected to a supply of a $220\, V$ is dissipating some power ${P_1}.$ The coil is cut into half and the two halves are connected in parallel. The heater now dissipates a power ${P_2}.$ The ratio of power ${P_1}\,\,:\,\,{P_2}$ is

A$2:1$
B$1:2$
C$1:4$
D$4:1$

Medium

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

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