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Alenkinab [10]
4 years ago
13

The resistor potential is constant and the inductor emf increases. The resistor potential is constant and the inductor emf decre

ases. The resistor potential is constant and the inductor emf is constant. The resistor potential decreases and the inductor emf is constant. The resistor potential decreases and the inductor emf increases. The resistor potential decreases and the inductor emf decreases. The resistor potential increases and the inductor emf increases. The resistor potential increases and the inductor emf is constant. The resistor potential increases and the inductor emf decreases.

Physics
1 answer:
maks197457 [2]4 years ago
6 0

The question is incomplete, the complete question is

When an RL circuit is connected to a battery, what happens to the potential difference across the resistor and the emf across the inductor?

The resistor potential is constant and the inductor emf increases.

The resistor potential is constant and the inductor emf decreases.

The resistor potential is constant and the inductor emf is constant.

The resistor potential decreases and the inductor emf is constant.

The resistor potential decreases and the inductor emf increases.

The resistor potential decreases and the inductor emf decreases.

The resistor potential increases and the inductor emf increases.

The resistor potential increases and the inductor emf is constant.

The resistor potential increases and the inductor emf decreases.

Answer:

The resistor potential increases and the inductor emf decreases.

Explanation:

From Kirchoff's rule, we can easily see that voltage across the inductor decreases steadily in a RL circuit until it finally gets to zero when the circuit is connected to a battery. A graph of the drop in potential across the inductor is attached for more clarity.

The drop shown in figure (b) in the image attached is the drop in potential across the inductor when an RL circuit is connected to a battery.

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b).

If the pulsar will continue to decelerate at this rate, it will  stop rotating at time:

t=\frac{w}{a_p}

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t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s

t=1016298.8 years

c).

582 years ago to 2019

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T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s

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Answer:

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