Question

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A domestic lighting circuit has a fuse of 5 A. If the mains supply is at 230 V, calculate the maximum number of 36 W tube-lights that can be safely used in this circuit?


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Answer 1

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Given data:

• Voltage supply = 230 V
• Current passing through circuit = 5 A
• Power consumed by one tube light = 36 W
• Let the total number of tube-light  = n

Solution:

Total power consumed = 36 × n W

Now, we have the formula

Power = Voltage × Current

Put all the values in this formula, we get

36 × n = 230 × 5

⇒ n = 31.99

Since the number of tube-light can not be in fraction,

$\boxed {\tt{hence \: \: n = 31}}$

Therefore, the maximum number of 36 W tube-lights that can be safely used in this circuit = 31

Answer 2

Let the maximum number of tube-lights be y.

Power of y− tube-lights, P=36y

V=230V,I=5A

We know that

P=VI

36y=230×5

36y=1500

y=31.94

So, number of tube-lights required are 31.

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