Video: Analyzing a Resistive- Capacitive Series Circuit

A 500-Ξ© resistor, an uncharged 1.50-ΞΌF capacitor, and a 6.16-V emf are connected in series. What is the initial current? What is the 𝑅𝐢 time constant? What is the current after one time constant? What is the voltage on the capacitor after one time constant?

05:15

Video Transcript

A 500-ohm resistor, an uncharged 1.50-microfarad capacitor, and a 6.16-volt emf are connected in series. What is the initial current? What is the 𝑅𝐢 time constant? What is the current after one time constant? What is the voltage on the capacitor after one time constant?

We can call our resistor, 500 ohms, capital 𝑅. And we can name our capacitor value, 1.50 microfarads, 𝐢. The 6.16- volt emf we can call capital 𝑉.

In part one, we want to know what is the initial current, what we’ll call 𝐼 sub 𝑖. In part two, we want to solve for the 𝑅𝐢 time constant, what we’ll call 𝜏. Next, we wanna solve for the current after one time constant has elapsed. We’ll call this 𝐼 sub one. And finally, we want to solve for the capacitor voltage after one time constant has elapsed. We’ll call this 𝑉 sub one.

We can start by considering a sketch of the circuit. In a powered 𝑅𝐢 circuit, like the one shown here, we have three important elements: the power supply, 𝑉; the resistor, 𝑅; and the capacitor, 𝐢. We’re told the capacitor is initially uncharged. And we want to solve at first for the initial current that’s in the circuit. At that initial moment, before the capacitor accumulates any charge, the current in the circuit follows Ohm’s law.

We can write that 𝐼 sub 𝑖 equal 𝑉 divided by 𝑅, essentially ignoring the capacitor for this moment in time. When we enter the given values for 𝑉 and 𝑅 and calculate this fraction, we find that 𝐼 sub 𝑖 is 12.3 milliamps. That’s the current in the circuit at time 𝑑 equals zero. Next, we wanna solve for the time constant we’ve called 𝜏. 𝜏 is equal to the resistance of our circuit 𝑅 multiplied by its capacitance 𝐢. 𝜏 equals 𝑅 times 𝐢, both of which are given to us in the problem statement.

So we can now plug in to solve for 𝜏. 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. So multiplying units of ohms times farads results in units of seconds. Next, we want to solve for the current in our circuit after one time constant has elapsed.

Because we have an 𝑅𝐢 circuit, there is a special mathematical relationship not just for current as a function of time but also voltage, which we’ll write down now in preparation for the last step. In an 𝑅𝐢 circuit, current as a function of time is equal to the initial current in the system 𝐼 sub 𝑖 multiplied by 𝑒 to the negative 𝑑 over 𝑅𝐢, where 𝑑 is the time elapsed. Similarly, the voltage across the capacitor, 𝑉, as a function of 𝑑 is equal to the initial voltage multiplied by one minus 𝑒 to the negative 𝑑 over 𝑅𝐢.

And remember that 𝑅𝐢 is also known as 𝜏, the time constant. We want to solve for current after one time constant has elapsed. 𝐼 sub one is equal to the initial current in the system 𝐼 sub 𝑖 times 𝑒 to the negative 𝜏 divided by 𝜏, where again 𝜏 is equal to 𝑅 times 𝐢. So the exponent is equal to negative one. And our equation simplifies to 𝐼 sub one equals 𝐼 sub 𝑖 over 𝑒.

We’ve solved for 𝐼 sub 𝑖 in a previous part. And 𝑒 is a natural number. Plugging in for 𝐼 sub 𝑖, 12.3 times 10 to the negative third amperes, and dividing it by 𝑒, which is approximately 2.78, we find a value for 𝐼 sub one of 4.53 milliamps. That’s how much current is running through the circuit after the capacitor has been charging for one time constant.

Finally, we want to solve for the voltage across the capacitor after one time constant has elapsed. To do that, we’ll use our 𝑉 as a function of 𝑑 equation for 𝑅𝐢 circuits. 𝑉 sub one is equal to 𝑉, the initial emf in our circuit, multiplied by one minus 𝑒 to the negative 𝜏 over 𝜏. So we see that the exponent once again goes to negative one. And our equation simplifies to 𝑉 times one minus one over 𝑒.

Entering in our value for 𝑉, 6.16 volts, and multiplying these two terms on our calculator, we find that 𝑉 sub one, to three significant figures, is equal to 3.89 volts. That’s the potential difference across the capacitor after its been charging for one time constant.

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