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Karolina [17]
3 years ago
14

A TV with a power rating of 267 W uses 1.9 kWh in one day. How many hours was the TV during that day?

Physics
1 answer:
Aleonysh [2.5K]3 years ago
4 0
1.9kWh/0.267kW = 7.11 hours
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A resistor with r = 340 ω and an inductor are connected in series across an ac source that has voltage amplitude 490 v. The rate
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The value of impedance Z of the circuit, when the rate at which electrical energy is dissipated in the resistor is 316 w, is 508 ohms.

<h3>What is impedance Z of the circuit?</h3>

The impedance Z of the circuit is the ratio of voltage amplitude to the maximum current.

Z=\dfrac{V}{I}

Here, <em>V </em>is voltage amplitude and<em> I</em> maximum current.

A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 490V. The rate at which electrical energy is dissipated in the resistor is 316 W.

The rate at which electrical energy is dissipated in the resistor is the product of the resistance and the square of current. Thus,

316=340\times I^2\\I=\sqrt{\dfrac{316}{340}}\\I=0.964\rm\; A

The impedance Z of the circuit is,

Z=\dfrac{V}{I}\\Z=\dfrac{490}{0.964}\\Z=508\rm\; ohm

Thus, the value of impedance Z of the circuit, when the rate at which electrical energy is dissipated in the resistor is 316 w, is 508 ohms.

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