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A series LR circuit consists of a 2.0-H inductor with negligible internal resistance, a 100-ohm resistor, an open switch, and a 9.0-V ideal power source. After the switch is closed, what is the maximum power delivered by the power supply?

Respuesta :

Answer:

The maximum power delivered by the power supply is 0.81 W.

Explanation:

Given that,

Inductance L= 2.0 H

Resistance R = 100 ohm

Voltage = 9.0 V

We need to calculate the power

Using formula of power

[tex]P = \dfrac{V^2}{R}[/tex]

Where, P = power

V = voltage

R = resistance

Put the value into the formula

[tex]P = \dfrac{(9.0)^2}{100}[/tex]

[tex]P =0.81\ W[/tex]

Hence, The maximum power delivered by the power supply is 0.81 W.

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