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3 years ago

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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?

1 answer:

VMariaS [17]3 years ago

7 0

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.

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<u>Explanation</u>:

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$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

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$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

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The number of turns in the secondary winding are 67.8 turns.

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