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hoa [83]
3 years ago
5

If a 60 W light bulb and a 75 W light bulb operate from 150 V source, which bulb has a greater current in it?

Physics
1 answer:
Monica [59]3 years ago
8 0

Answer:

Since it us the resistance that causes the power output, it might be intuitive to think that more resistance produces more power but as P=I^2 * R one actually needs high current and low resistqnce. That means 75 W has lower resistance and higher current.

Now you may think that making the resistance very low, one gets the most power. Not so, as the circuit has wires, power source etc. All the have their resistance. In theory one gets the maximum power over a resistor when the resistor is half of the total resistance of the circuit. This in practice would be inefficient as one would lose half of the total power elsewhere. In practice circuits have fuses that limit the current and one gets the maximum power at the current the fuse is rated to. So if there is 20 A fuse, the maximum power is 2400 W and the corresponding resistance 6 ohm.

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Answer:

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3 0
2 years ago
A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

8 0
3 years ago
คลื่นกลเกิดขึ้นได้อย่างไร
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I hope it helps, Regards.

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