Question

In: Physics

1. When an RL circuit is connected to a battery, what happens to the potential difference...

1. When an RL circuit is connected to a battery, what happens to the potential difference across the inductor?

It is constant.

It is initially at its maximum value and then decreases exponentially with time.

It is initially at its maximum value and then decreases linearly with time.

It is initially 0 and then increases.

2. If we decrease the resistance in an RL circuit, what happens to the time required for the current to reach, say, 50% of its final value after the battery is connected?

The time is increased.

The time is decreased.

The time is the same.

3. In the two-coil setup in this section, if the current in coil 1 is increasing at a constant rate, which is true?

The induced emf in coil 2 is increasing.

The induced emf in coil 2 is decreasing.

The induced emf in coil 2 is constant.

4. In the two-coil setup in this section, if the current in coil 1 is decreasing at a constant rate, which is true?

The induced emf in coil 2 is constant.

The induced emf in coil 2 is increasing.

The induced emf in coil 2 is decreasing.

Solutions

Expert Solution

1. Voltage across inductor is at its maximum value and then decreases exponentially with time. As we know that the expression of voltage across inductor can be given as :-

; where v(t) is the voltage across inductor and V is source voltage and T is time constant. As the expression is exponential and decreases with time, so the second option is correct.

2. The equation for the time to 50 % of the final value is given as :-

; where t is the time for current to reach 50% of final value.

Clearly, when the value of R is decreased time for the current to reach 50% of final value will increase.

3. The induced emf for the second coil can be given in terms of mutual inductance as follows :-

e = -M21 (dI1/dt); where M21 is the mutual inductance

Here, rate of change of current in coil 1 is constant, so induced emf will also be constant. Hence, option 3 is true.

4. As per question number 3, same will be case as current is decreasing only, but rate is constant. So the enduced emf will be constant, but with reversed polarity compared to 3 case. Hence, option 1 is correct.

genius_generous answered 1 year ago

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