A circuit consists of a coil, a switch, and a battery, all...

A circuit consists of a coil, a switch, and a battery, all in series. The internal resistance of the battery is negligible compared with that of the coil. The switch is originally open. It is thrown closed, and after a time interval $\Delta t,$ the current in the circuit reaches $80.0 \%$ of its final value. The switch then remains closed for a time interval much longer than $\Delta t$. The wires connected to the terminals of the bat- tery are then short-circuited with another wire and removed from the battery, so that the current is uninterrupted. (a) At an instant that is a time interval $\Delta t$ after the short circuit, the current is what percentage of its maximum value? (b) At the moment $2 \Delta t$ after the coil is short-circuited, the current in the coil is what percentage of its maximum value?

A circuit consists of a coil, a switch, and a battery, all in series. The internal resistance of the battery is negligible compared with that of the coil. The switch is originally open. It is thrown closed, and after a time interval $\Delta t,$ the current in the circuit reaches $80.0 \%$ of its final value. The switch then remains closed for a time interval much longer than $\Delta t$. The wires connected to the terminals of the bat-
tery are then short-circuited with another wire and removed from the battery, so that the current is uninterrupted. (a) At an instant that is a time interval $\Delta t$ after the short circuit, the current is what percentage of its maximum value? (b) At the moment $2 \Delta t$ after the coil is short-circuited, the current in the coil is what percentage of its maximum value?

Key Concepts

Exponential Growth and Decay

In RL circuits, both the rising and falling currents follow an exponential behavior. When a voltage is applied, the current increases according to an exponential growth function until it approaches its maximum steady-state value. Conversely, when the circuit is disconnected from the battery and only a resistive pathway remains (such as in a short circuit), the stored magnetic energy in the inductor causes the current to decay exponentially over time.

Inductor Current Continuity

One key property of an inductor is that its current cannot change instantaneously. When circuit conditions change abruptly, such as during a short-circuiting process, the inductor forces the current to remain continuous, resulting in exponential decay rather than an immediate drop to zero. This principle is crucial in understanding transient responses in circuits involving inductors.

RL Circuit Transient Response

This concept describes how the current in a resistor–inductor circuit builds up or decays exponentially over time. When the circuit is closed or opened, the current does not instantly reach its new steady-state value but instead changes gradually following an exponential law governed by the properties of the resistor and the inductor.

Time Constant (? = L/R)

The time constant is a parameter defined as the ratio of the inductance L to the resistance R, and it characterizes the rate at which current builds up or decays in an RL circuit. A time interval equal to the time constant corresponds to the point where the current has reached about 63.2% of its final steady-state value in a growing current scenario, or has fallen to about 36.8% of its initial value in a decaying current scenario.

Transcript

00:02 Hi there, so for this problem, we know that with the information that we are given, first of all, we consider that for the increased current, we know that the current is equal to the mth, this divided by the resistance, and this times one minus the exponential of minus the inductance times the time divided by the resistance.

00:34 Now the final value for this, we know that the final value for this is the nph divided by the resistance.

00:42 So the condition on the interval of time, delta t, is that after that time, this is going to be equal to 0 .8 times the value of the nph divided by the resistance.

00:59 So we just equal this to expression that we are given.

01:03 So as you can see, we can simplify terms in here.

01:11 We can cancel this with this.

01:14 Then solving in there, we will have that the exponential.

01:18 What we have done is to simply pass this term to the other side and then subtract 1 minus 0 .8.

01:28 So we obtain the following.

01:32 We obtain that this is equal to 0 .2.

01:38 Then what we can do in this case is to invert and we can take the inverse in both sides of this expression.

01:50 So we obtain the s penitial.

01:52 Now with a positive argument, so that will be the inductance times the interval of time, this divided by the resistance.

02:01 Then this is equal to five.

02:06 Now with that, we can apply the nefarion logarithm.

02:11 In both sides of this expression, so we will have the inductance times the interval of time, and this divided by the resistance...

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