Two electric bulbs rated {P_1},watt V, volts and {P_2}, watt V, volts are connected in parallel and V, volts are applied to it. The total

Q: Two electric bulbs rated ${P_1}\,watt$ $V\, volts$ and ${P_2}\, watt$ $V\, volts$ are connected in parallel and $V\, volts$ are applied to it. The total power will be

a If resistances of bulbs are ${R_1}$ and ${R_2}$ respectively then in parallel $\frac{1}{{{R_P}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}$$ \Rightarrow $ $\frac{1}{{\left( {\frac{{{V^2}}}{{{P_p}}}} \right)}} = \frac{1}{{\left( {\frac{{{V^2}}}{{{P_1}}}} \right)}} + \frac{1}{{\left( {\frac{{{V^2}}}{{{P_2}}}} \right)}}$ $ \Rightarrow $ ${P_P} = {P_1} + {P_2}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

Two electric bulbs rated ${P_1}\,watt$ $V\, volts$ and ${P_2}\, watt$ $V\, volts$ are connected in parallel and $V\, volts$ are applied to it. The total power will be

A${P_1} + {P_2}\,watt$
B$\sqrt {{P_1}{P_2}}\, watt$
C$\frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}}\,watt$
D$\frac{{{P_1} + {P_2}}}{{{P_1}{P_2}}}\,watt$

Medium

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

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