Video: Calculating the Maximum Current in an RLC Circuit

A 6 ฮฉ resistor, a capacitor with reactance of 80 ฮฉ, and an inductor with negligible resistance and a self-inductance coefficient of 0.28 H are connected in series with an AC power supply of 20 V at 50 Hz. Calculate the maximum current in the circuit, treating ๐œ‹ as exactly 3.14.

04:07

Video Transcript

A six-ohm resistor, a capacitor with reactance of 80 ohms, and an inductor with negligible resistance and a self-inductance coefficient of 0.28 henries are connected in series with an AC power supply of 20 volts at 50 hertz. Calculate the maximum current in the circuit, treating ๐œ‹ as exactly 3.14.

Letโ€™s start out by sketching this AC circuit. The elements in this electrical circuit are an AC power supply, a resistor, a capacitor, and an inductor. Weโ€™re given some information about each of these elements and we want to calculate the maximum current that ever runs through this circuit.

To do that, letโ€™s consider a relationship between current, voltage, and the overall resistance of this circuit. The maximum current ever running through a circuit like this is equal to the maximum voltage supplied by the power source divided by the impedance ๐‘ of the circuit. This impedance is analogous to resistance. And it comes up and we consider AC-power-supplied circuits.

In general, impedance has three components that can contribute to it. The first is the resistance of the circuit, the second is the inductive reactance ๐‘‹ sub ๐ฟ, and the third is the capacitive reactance ๐‘‹ sub ๐ถ of the circuit.

Based on our equation for ๐ผ sub max, which is what we want to solve for, we know that to do that, weโ€™ll have to solve for the maximum voltage supplied as well as the impedance in the circuit. Letโ€™s start with the maximum voltage ๐‘‰ sub max.

In our problem statement, weโ€™re given the voltage of our power supply at 20 volts. However, this is not equal to ๐‘‰ sub max. When the voltage supplied to a circuit varies as it does in this case with an AC power supply, itโ€™s common to give the voltage of that power supply at a level thatโ€™s known as the effective voltage ๐‘‰ sub eff. This effective voltage is neither the highest nor the lowest voltage supplied by the sinusoidal power supply.

So ๐‘‰ sub eff and ๐‘‰ sub max are not the same thing. But thankfully, there is an equation that does relate these two. If we take our given value for the effective voltage and multiply it by the square root of two over two, then that value will be equal to the maximum voltage output by the power supply.

If we plug in 20 volts for the effective voltage and calculate, we find that the maximum voltage is equal to 28.2885 volts. Thatโ€™s the most voltage that our power supply will ever create.

Now that we know the numerator of the fraction which will tell us the maximum current, weโ€™ll look to solve for the denominator, the impedance of our circuit. In our problem statement, weโ€™re told that the resistor in this circuit has a resistance of six ohms and that the capacitor has a reactance of 80 ohms.

In other words, weโ€™ve just been told capital ๐‘… and ๐‘‹ sub ๐ถ for this circuit. However, weโ€™re never told ๐‘‹ sub ๐ฟ, the inductive reactance. To solve for that term, weโ€™ll need to use the fact that ๐‘‹ sub ๐ฟ is equal to two times ๐œ‹ times the frequency of oscillation of the circuit multiplied by the self-inductance coefficient capital ๐ฟ.

The frequency of the circuit ๐‘“ is given as 50 hertz and capital ๐ฟ is given as 0.28 henries. Treating ๐œ‹ as exactly 3.14, when we multiply all these numbers together. We find that ๐‘‹ sub ๐ฟ, the inductive reactance, is 87.92 ohms.

Now that we know that value, we have all the components that go into making up the impedance of the circuit ๐‘. When we plug in for these three values, we use six ohms for our resistance ๐‘… 87.92 ohms for inductive reactance and 80 ohms given in our problem statement as the capacitive reactance. Combining these terms this way gives us an overall circuit impedance of 9.936 ohms.

What weโ€™ve now accomplished is weโ€™ve solved for the numerator and denominator that weโ€™ll need in order to calculate the maximum current in the circuit. That maximum current is equal to 28.2885 volts divided by 9.936 ohms. That value is equal to 2.847 amps. Thatโ€™s the maximum current in this circuit.

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