WEBVTT
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A 500-ohm resistor, an uncharged 1.50-microfarad capacitor, and a 6.16-volt emf are connected in series.
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What is the initial current?
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What is the π
πΆ time constant?
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What is the current after one time constant?
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What is the voltage on the capacitor after one time constant?
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We can call our resistor, 500 ohms, capital π
.
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And we can name our capacitor value, 1.50 microfarads, πΆ.
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The 6.16- volt emf we can call capital π.
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In part one, we want to know what is the initial current, what weβll call πΌ sub π.
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In part two, we want to solve for the π
πΆ time constant, what weβll call π.
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Next, we wanna solve for the current after one time constant has elapsed.
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Weβll call this πΌ sub one.
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And finally, we want to solve for the capacitor voltage after one time constant has elapsed.
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Weβll call this π sub one.
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We can start by considering a sketch of the circuit.
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In a powered π
πΆ circuit, like the one shown here, we have three important elements: the power supply, π; the resistor, π
; and the capacitor, πΆ.
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Weβre told the capacitor is initially uncharged.
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And we want to solve at first for the initial current thatβs in the circuit.
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At that initial moment, before the capacitor accumulates any charge, the current in the circuit follows Ohmβs law.
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We can write that πΌ sub π equal π divided by π
, essentially ignoring the capacitor for this moment in time.
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When we enter the given values for π and π
and calculate this fraction, we find that πΌ sub π is 12.3 milliamps.
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Thatβs the current in the circuit at time π‘ equals zero.
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Next, we wanna solve for the time constant weβve called π.
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π is equal to the resistance of our circuit π
multiplied by its capacitance πΆ.
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π equals π
times πΆ, both of which are given to us in the problem statement.
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So we can now plug in to solve for π.
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And when we do, being careful to use units of farads for our capacitance, and multiply these two numbers together, we find that our time constant is 7.50 times 10 to the negative fourth seconds.
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So multiplying units of ohms times farads results in units of seconds.
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Next, we want to solve for the current in our circuit after one time constant has elapsed.
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Because we have an π
πΆ circuit, there is a special mathematical relationship not just for current as a function of time but also voltage, which weβll write down now in preparation for the last step.
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In an π
πΆ circuit, current as a function of time is equal to the initial current in the system πΌ sub π multiplied by π to the negative π‘ over π
πΆ, where π‘ is the time elapsed.
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Similarly, the voltage across the capacitor, π, as a function of π‘ is equal to the initial voltage multiplied by one minus π to the negative π‘ over π
πΆ.
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And remember that π
πΆ is also known as π, the time constant.
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We want to solve for current after one time constant has elapsed.
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πΌ sub one is equal to the initial current in the system πΌ sub π times π to the negative π divided by π, where again π is equal to π
times πΆ.
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So the exponent is equal to negative one.
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And our equation simplifies to πΌ sub one equals πΌ sub π over π.
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Weβve solved for πΌ sub π in a previous part.
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And π is a natural number.
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Plugging in for πΌ sub π, 12.3 times 10 to the negative third amperes, and dividing it by π, which is approximately 2.78, we find a value for πΌ sub one of 4.53 milliamps.
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Thatβs how much current is running through the circuit after the capacitor has been charging for one time constant.
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Finally, we want to solve for the voltage across the capacitor after one time constant has elapsed.
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To do that, weβll use our π as a function of π‘ equation for π
πΆ circuits.
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π sub one is equal to π, the initial emf in our circuit, multiplied by one minus π to the negative π over π.
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So we see that the exponent once again goes to negative one.
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And our equation simplifies to π times one minus one over π.
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Entering in our value for π, 6.16 volts, and multiplying these two terms on our calculator, we find that π sub one, to three significant figures, is equal to 3.89 volts.
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Thatβs the potential difference across the capacitor after its been charging for one time constant.