Video: Finding the Current in an RL Circuit

How long after switch 𝑆₁ is thrown does it take the current in the circuit shown to reach half its maximum value? Express your answer in terms of the time constant of the circuit.

02:55

Video Transcript

How long after switch 𝑆 one is thrown does it take the current in the circuit shown to reach half its maximum value? Express your answer in terms of the time constant of the circuit.

In this exercise statement, we want to solve for a time value: how long after the switch is closed that the current reduced half its maximum value? We’ll call this time 𝑑 sub one-half. When we look at the circuit, we see it has a power supply, a resistor 𝑅, and an inductor 𝐿.

With the switch 𝑆 one open, no current flows throughout the circuit. But as soon as it closes, current will begin to flow and ramp up to its maximum value. It will instantly reach its maximum value thanks to the inductor that’s part of this circuit.

If we name this circuit an 𝑅𝐿 circuit because of the components involved, then the current that runs through the circuit is given as a function of time after the switch is closed. Current as a function of time in the circuit once the switch is closed, representing time 𝑑 equals zero, is equal to the power supplied divided by the resistance multiplied by one minus 𝑒 to the negative 𝑑 over 𝜏, where 𝜏 is the time constant of the circuit.

When we apply this relationship to our scenario, notice that the factor πœ€ over 𝑅, representing the voltage over the resistance, is equal by Ohm’s law to the maximum current that runs through this circuit. We’ll call it 𝐼 sub max. So we can say that the current in the circuit at any time 𝑑 is equal to the maximum current that can run through the circuit multiplied by this term in parentheses.

At a particular time, when 𝑑 is equal to 𝑑 one-half, we can write that 𝐼 sub max over two, half the maximum current, is equal to 𝐼 sub max multiplied by one minus 𝑒 to the negative 𝑑 sub one-half over 𝜏.

We can see here that the maximum current 𝐼 sub max cancels out from both sides. And we find that one-half equals one minus 𝑒 to the negative 𝑑 sub one-half over 𝜏. Or 𝑒 to the negative 𝑑 sub one-half over 𝜏 is equal to one-half. And if we take the natural logarithm of both sides, then we can rewrite the left-hand side of this equation as negative 𝑑 sub one-half over 𝜏.

We then multiply both sides by negative 𝜏. 𝑑 sub one-half equals negative 𝜏 times the natural log of one-half. When we evaluate the natural log of one-half, we find that, to two significant figures, it equals negative 0.69. So the time it takes for the current in the circuit to reach half its maximum value is equal to 0.69 times the time constant of the circuit.

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