Question

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?

Answer 1

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

$P = \dfrac{V^2}{R}$

Where, P = power

V = voltage

R = resistance

Put the value into the formula

$P = \dfrac{(9.0)^2}{100}$

$P =0.81\ W$

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