Two bulbs A and B are connected to a $220$ V supply. The maximum current that can flow through A and B is $0.5$ A and $1.2$ A respectively. The effective resistance of the two bulbs in series and parallel is $283.3$ $Ω$ and $58.6$ $Ω$ respectively. the current flowing through the circuit when they are connected in series is _ A and in parallel is _A.

0.78, 3.7

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2.4, 1.2

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3.6, 8.6

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0.2, 1.7

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Solution

The correct option is A $0.78, 3.7$
Given:-
$R_{1} + R_{2} = 283.3 \to (1) \frac{R_{1} R_{2}}{R_{1} + R_{2}} = 58.6 \to (2)$
pit in (2)
$\frac{R_{1} R_{2}}{283.3} = 58.6 R_{1} R_{2} = 283.3 \times 58.6 R_{2} = \frac{283.3 \times 58.6}{R_{1}}$
when in series ,current $= \frac{V}{R_{n e t}}$
$= \frac{220}{283.3} = 0.78 A$
when in parallel ,current $= \frac{V}{R_{n e t}}$
$= \frac{220}{58.6} = 3.7 A$

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