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3 years ago

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A stationary conductive loop with an internal resistance of 0.5 Ω is placed in a time varying magnetic field. When the loop is c

losed, a current of 5 A flows through it. What will the current be if the loop is opened to create a small gap and a 2-Ω resistor is connected across its open ends?

1 answer:

Setler [38]3 years ago

5 0

Answer:

e= -AdB/dt

5*0.5 = -AdB/dt

i(2+0.5) = -AdB/dt

from above expressions

2.5 = i2.5

so i = 1A

More Answers:

v=0.5x5= 2.5

2.5/(2+0.5) = 1 A

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From this question we are told that

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$\beta = \frac{Q_c}{P_{in}}$

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$\beta_{Carnot } = \frac{T_c}{T_h - T_c}$

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$\beta_{Carnot} = 9.267\\$

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$\beta_{air} = 0.5 * 9.2667$

$= 4.6333$

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$Q_c = \beta P_{in}$

$= 4.6333 *2$

$9.2667kW$

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$Q_c = \r mc_p \Delta T$

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