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$B_{1}$ , $B_{2}$ and $B_{3}$ are three identical bulbs connected, as shown in Figure. When all three bulbs glow, a current of $3 A$ is recorded by ammeter A.

How much power is dissipated in the circuit when all three bulbs glow together?

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13.5 W

4.5 W

14.5 W

15 W

answer is A.

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Detailed Solution

The power is dissipated in the circuit when all three bulbs glow together is

13.5 W

.
The power dissipated when all three bulbs glow together (

P_{1} = P_{2} = P_{3}

)

P_{eq} = P_{1} + P_{2} + P_{3}

\Rightarrow P_{eq} = 3 P

Since the resistance of each arm is the same, the current is divided equally among them. Hence

1 A

will pass through each bulb equally.

I = 1 A

The power

P

is given by the equation

P = V \times I

Where

V

is the voltage

\Rightarrow P_{eq} = 3 \times V \times I

\Rightarrow P_{eq} = 3 \times 4.5 \times 1

\Rightarrow P_{eq} = 13.5 W

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