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How long after switch π one is thrown does it take the current in the circuit shown to reach half its maximum value?
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Express your answer in terms of the time constant of the circuit.
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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?
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Weβll call this time π‘ sub one-half.
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When we look at the circuit, we see it has a power supply, a resistor π
, and an inductor πΏ.
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With the switch π one open, no current flows throughout the circuit.
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But as soon as it closes, current will begin to flow and ramp up to its maximum value.
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It will instantly reach its maximum value thanks to the inductor thatβs part of this circuit.
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If we name this circuit an π
πΏ 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.
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Current as a function of time in the circuit once the switch is closed, representing time π‘ equals zero, is equal to the power supplied divided by the resistance multiplied by one minus π to the negative π‘ over π, where π is the time constant of the circuit.
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When we apply this relationship to our scenario, notice that the factor π over π
, representing the voltage over the resistance, is equal by Ohmβs law to the maximum current that runs through this circuit.
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Weβll call it πΌ sub max.
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So we can say that the current in the circuit at any time π‘ is equal to the maximum current that can run through the circuit multiplied by this term in parentheses.
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At a particular time, when π‘ is equal to π‘ one-half, we can write that πΌ sub max over two, half the maximum current, is equal to πΌ sub max multiplied by one minus π to the negative π‘ sub one-half over π.
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We can see here that the maximum current πΌ sub max cancels out from both sides.
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And we find that one-half equals one minus π to the negative π‘ sub one-half over π.
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Or π to the negative π‘ sub one-half over π is equal to one-half.
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And if we take the natural logarithm of both sides, then we can rewrite the left-hand side of this equation as negative π‘ sub one-half over π.
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We then multiply both sides by negative π.
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π‘ sub one-half equals negative π times the natural log of one-half.
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When we evaluate the natural log of one-half, we find that, to two significant figures, it equals negative 0.69.
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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.